Chapter 08 - Sequences and Series

Sequences

Videos

Review Videos

Videos, Math Dr. Bob:

• Infinite sequences: Definition; Squeeze Theorem
  • Extra: Infinite sequences: Various examples, arithmetic and geometric
  • Extra: Infinite sequences: Recursive sequences (like Fibonacci)

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Theory

Theory 1

A sequence is a rule that defines a term for each natural number $n\in \mathbb{N}$:

$$a_0,a_1,a_2,a_3,a_4,\dots$$

So a sequence is a function from $\mathbb{N}$ to $\mathbb{R}$.

Geometric sequence

A sequence is called geometric if the ratio of consecutive terms is some constant $r$, independent of $n$:

$$\frac{a_{n+1}}{a_n}=r\text{for every}n$$

The defining relation of a geometric sequence is equivalent to $a_{n+1}=a_n\cdot r$.

By plugging $a_1=a_0\cdot r$ into $a_2=a_1\cdot r$, we have $a_2=(a_0\cdot r)\cdot r=a_0\cdot r^2$. This plugging can be repeated $n$-times to get a formula for the $n^{\text{th}}$ term:

$$a_n=a_{n-1}\cdot r=a_{n-2}\cdot r^2=a_{n-3}\cdot r^3=\cdots =a_1\cdot r^{n-1}=a_0\cdot r^n$$

Therefore $a_n=a_0\cdot r^n$, and we have a formula for the general term of the sequence (the term with index $n$).

Starting point of a sequence

Note that sometimes the index (variable) of a sequence starts somewhere other than $0$. Most common is $1$ but any other starting point is allowed, even negative numbers.

Sometimes $c$ is used instead of $a_0$ in the formula for the general term of a sequence, thus $a_n=c{r}^n$. The ‘$c$’ notation is useful when the sequence starts from $n\ne 0$.

Extra - Fibonacci sequence

The Fibonacci sequence goes like this:

$$0,1,1,2,3,5,8,13,21,34,55,89,144,\dots$$

The pattern is:

$$F_n=F_{n-1}+F_{n-2}$$

This formula is a recursion relation, which means that terms are defined using the values of prior terms.

The Fibonacci sequence is perhaps the most famous sequence of all time. It is related to the Golden Ratio and the Golden Spiral:

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Illustration

Example: Geometric sequence - revealing the format

Geometric sequence: revealing the format

Find $a_0$ and $r$ and $a_n$ (written in the geometric sequence format) for the following geometric sequences:

(a) $a_n={\left(-\frac{1}{2}\right)}^n$
(b) $b_n=-3\left(\frac{2^{n+1}}{5^n}\right)$
(c) $c_n=e^{5-7n}$

Solution

(a)

Plug in $n=0$ to obtain $a_0=1$. Notice that $a_{n+1}/a_n=-1/2$ and so therefore $r=-1/2$. Then the ‘general term’ is $a_n=a_0\cdot r^n=1\cdot (-1/2{)}^n$.

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(b)

Rewrite the fraction:

$$\frac{2^{n+1}}{5^n}\gg \gg 2\cdot {\left(\frac{2}{5}\right)}^n$$

Plug that in and observe $b_n=-6\cdot (2/5{)}^n$. From this format we can read off $b_0=-6$ and $r=2/5$.

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(c)

Rewrite:

$$c_n\gg \gg e^5\cdot e^{-7n}\gg \gg e^5\cdot {\left(e^{-7}\right)}^n$$

From this format we can read off $c_0=e^5$ and $r=e^{-7}$.

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Series

Videos

Review Videos

Videos, Math Dr. Bob:

• Infinite series: Definitions, basic examples
• Geometric series and SDT: Geometric series, Simple Divergence Test (aka “Limit Test”)
• Infinite series: Various examples
  • Extra: Infinite series convergence: Telescoping series

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Theory

Theory 1

A series is an infinite sum that is created by successive additions without end. The terms are not added up “all at once” but rather they are added up “as $n$ increases” or “as $n\to \infty$.”

$$a_0+a_1+a_2+a_3+\dots =\sum _{n=0}^{\infty }{a}_n$$

Three of the most famous series are the Leibniz series and the geometric series:

$$\begin{array}{rlrl}\text{Leibniz series:}& 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots +\frac{(-1{)}^n}{2n+1}+\cdots & =& \frac{\pi }{4}\\ & & & \\ \text{Geometric series:}& 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +{\left(\frac{1}{2}\right)}^n+\cdots & =& 2\end{array}$$

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Partial sum sequence of a series

The partial sum sequence of a series is the sequence whose terms are the sums up to the given index:

$$S_N=a_0+a_1+\cdots +a_N=\sum _{n=0}^N{a}_n$$

These $S_N$ terms themselves form a sequence:

$$S_0,S_1,S_2,S_3,\dots$$

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Illustration

Example: Geometric series - total sum and partial sums

Geometric series - total sum and partial sums

The geometric series total sum $S$ can be calculated using a “shift technique” as follows:
(1) Compare $S$ and $rS$:

$$\begin{array}{rl}S=& a_0+a_{0}r+a_0{r}^2+a_0{r}^3+\cdots \\ \stackrel{\times r}{\gg \gg }rS=& a_{0}r+a_0{r}^2+a_0{r}^3+a_0{r}^4+\cdots \end{array}$$

(2) Subtract second line from first line, many cancellations:

$$\begin{array}{rl}S& =a_0+a_{0}r+a_0{r}^2+a_0{r}^3+\cdots \\ -(rS& =a_{0}r+a_0{r}^2+a_0{r}^3+a_0{r}^4+\cdots )\\ ———& ———————————————\\ S-rS& =a_0\end{array}$$

(3) Solve to find $S$:

$$S=\frac{a_0}{1-r}$$

Assumes convergence!

Note: this calculation assumes that $S$ exists, i.e. that the series converges.

The geometric series partial sums can be calculated similarly, as follows:

(1) Compare $S$ and $rS$:

$$\begin{array}{rl}{S}_N=& a_0+a_{0}r+a_0{r}^2+\cdots +a_0{r}^N\\ \stackrel{\times r}{\gg \gg }r{S}_N=& a_{0}r+a_0{r}^2+\cdots +a_0{r}^N+a_0{r}^{N+1}\end{array}$$

(2) Subtract second line from first line, many cancellations:

$$\begin{array}{rl}{S}_N=& a_0+a_{0}r+a_0{r}^2+\cdots +a_0{r}^N\\ -(r{S}_N=& a_{0}r+a_0{r}^2+\cdots +a_0{r}^N+a_0{r}^{N+1})\\ ———& ———————————————\\ S_N-r{S}_N& =a_0-a_0{r}^{N+1}\end{array}$$

(3) Solve to find $S_N$:

$$\begin{array}{c}{S}_N=a_0\frac{1-r^{N+1}}{1-r}\\ \\ =\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{N+1}=S-S{r}^{N+1}\end{array}$$

(4) The last formula is revealing in its own way. Here is what it means in terms of terms:

$$\begin{array}{rl}{a}_0+a_{0}r+\cdots +a_0{r}^N=& \\ & \\ & \begin{array}{rl}{a}_0+a_{0}r+a_0{r}^2+\cdots & \\ & -(a_0{r}^{N+1}+a_0{r}^{N+2}+\cdots )\end{array}\end{array}$$Link to original

Convergence

Videos

Review Videos

Videos, Math Dr. Bob:

• Infinite sequences convergence: Squeeze; Monotone Bounded
• Infinite sequences convergence: Examples sequences: convergent, monotonic, bounded

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Theory

Theory 1

A sequence has a limit if its terms tend toward a specific number, or toward $\pm \infty$.

When this happens we can write “${\lim }_{n\to \infty }{a}_n=L$” with some number $L\in \mathbb{R}$ or $L=\pm \infty$.
We can also write “$a_n\to L$ as $n\to \infty$”.

The sequence is said to converge if it has a finite limit $L\in \mathbb{R}$.

Some sequences don’t have a limit at all, like $a_n=\cos n$:

Or $a_n=e^n$:

These sequences diverge.
In the second case, there is a limit $L=\infty$, so we say it diverges to $+\infty$.

A sequence may have a limit of $\pm \infty$ but is still said to diverge.

Extra - Convergence definition

The precise meaning of convergence is this. We have $a_n\to L$ as $n\to \infty$ if, given any proposed error $\epsilon >0$, it is possible to find $N$ such that for all $n>N$ we have $|a_n-L|<\epsilon$.

When $L=\infty$, convergence means that given any $B>0$, we can find $N$ such that for all $n>N$ we have $a_n>B$.

Similarly for $L=-\infty$.

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If the general term $a_n$ is a continuous function of $n$, we can replace $n$ with the continuous variable $x$ and compute the continuous limit instead:

$$\underset{n\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }{a}_x$$

If $a_x$ would be a differentiable function, and we discover an indeterminate form, then we can apply L’Hopital’s Rule to find the limit value. For example, if the indeterminate form is $0\cdot \infty$, we can convert it to $\frac{\infty }{1/0}=\frac{\infty }{\infty }$ and apply L’Hopital.

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Illustration

Example - L’Hopital’s Rule for sequence limits

L’Hopital’s Rule for sequence limits

(a) What is the limit of $a_n=\frac{\ln n}{n}$?
(b) What is the limit of $b_n=\frac{(\ln n{)}^2}{n}$?
(c) What is the limit of $c_n=n\left(\sqrt{n^2+1}-\sqrt{n^2}\right)$?
(d) What is the limit of $d_n=n{e}^{-n}$?

Solution

(a)

Identify indeterminate form $\frac{\infty }{\infty }$. Change from $n$ to $x$ and apply L’Hopital:

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x}\stackrel{\frac{d}{dx}}{\gg \gg }\underset{x\to \infty }{\lim }\frac{1/x}{1}=0$$

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(b)

Identify indeterminate form $\frac{\infty }{\infty }$. Change from $n$ to $x$ and apply L’Hopital:

$$\underset{x\to \infty }{\lim }\frac{(\ln x{)}^2}{x}\stackrel{\frac{d}{dx}}{\gg \gg }\underset{x\to \infty }{\lim }\frac{2\ln x\cdot \frac{1}{x}}{1}=2\underset{x\to \infty }{\lim }\frac{\ln x}{x}\stackrel{\text{(by (a) result)}}{=}0$$

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(c)

(1) Identify form $\infty \cdot 0$ and rewrite as $\frac{\infty }{\infty }$:

$$n\left(\sqrt{n^2+1}-\sqrt{n^2}\right)\gg \gg \frac{\sqrt{n^2+1}-\sqrt{n^2}}{1/n}$$

(2) Change from $n$ to $x$ and apply L’Hopital:

$$\frac{\sqrt{x^2+1}-\sqrt{x^2}}{1/x}\gg \gg \frac{\frac{1}{2}{\left(x^2+1\right)}^{-1/2}(2x)-1}{-1/x^2}$$

(3) Simplify:

$$\gg \gg \frac{-2x^3}{\sqrt{x^2+1}}+\frac{1}{x^2}\gg \gg \frac{-2x^5+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}$$

(4) Consider the limit:

$$\frac{-2x^5+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}\stackrel{x\to \infty }{⟶}\frac{-2x^5+x}{x^3}⟶-\infty$$

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(d)

(1) Identify form $\infty \cdot 0$ and rewrite as $\frac{\infty }{\infty }$:

$$n{e}^{-n}\gg \gg \frac{n}{e^n}$$

(2) Change $n$ to $x$ and apply L’Hopital:

$$\frac{x}{e^x}\gg \gg \frac{1}{e^n}\stackrel{x\to \infty }{⟶}0$$ Link to original

Example - Squeeze theorem

Squeeze theorem

Use the squeeze theorem to show that $\frac{4^n}{n!}\to 0$ as $n\to \infty$.

Solution

(1) We will squeeze the given general term above $0$ and below a sequence $b_n$ that we must devise:

$$0\le \frac{4^n}{n!}\le b_n$$

(2) We need $b_n$ to satisfy $b_n\to 0$ and $\frac{4^n}{n!}\le b_n$. Let us study $\frac{4^n}{n!}$.

$$\frac{4^n}{n!}=\frac{4\cdot 4\cdot \cdots \cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4}{n(n-1)\cdots 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$$

(3) Now for the trick. Collect factors in the middle bunch:

$$\frac{4^n}{n!}=\frac{4}{n}\left(\frac{4}{n-1}\cdot \frac{4}{n-2}\cdot \cdots \cdot \frac{4}{7}\cdot \frac{4}{6}\cdot \frac{4}{5}\right)\frac{4\cdot 4\cdot 4\cdot 4}{4\cdot 3\cdot 2\cdot 1}$$

(4) Each factor in the middle bunch is $<1$ so the entire middle bunch is $<1$. Therefore:

$$\frac{4^n}{n!}<\frac{4}{n}\cdot \frac{4^4}{4!}=\frac{1024}{24n}$$

Now we can easily see that $1024/24n\to 0$ as $n\to \infty$, so we set $b_n=1024/24n$ and we are done.

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Theory

Theory 2

Monotone sequences

A sequence is called monotone increasing if $a_{n+1}\ge a_n$ for every $n$.

A sequence is called monotone decreasing if $a_{n+1}\le a_n$ for every $n$.

In this context, ‘monotone’ just means it preserves the increasing or decreasing modality for all terms.

Monotonicity Theorem

If a sequence is monotone increasing, and bounded above by $B$, then it must converge to some limit $L$, and $L\le B$.

If a sequence is monotone decreasing, and bounded below by $B$, then it must converge to some limit $L$, and $L\ge B$.

Terminology:

• Bounded above by $B$ means that $a_n\le B$ for every $n$
• Bounded below by $B$ means that $B\le a_n$ for every $n$

Notice!

The Monotonicity Theorem says that a limit $L$ exists, but it does not provide the limit value.

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Illustration

Example - Monotonicity Theorem

Monotonicity

Show that $a_n=\sqrt{n+1}-\sqrt{n}$ converges.

Solution

(1) Observe that $a_n>0$ for all $n$.

Because $n+1>n$, we know $\sqrt{n+1}>\sqrt{n}$.

Therefore $\sqrt{n+1}-\sqrt{n}>0$

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(2) Show $a_x$ is decreasing:

Replace $n$ with $x$: $a_x=\sqrt{x+1}-\sqrt{x}$ considered as a differentiable function.

Take derivative to show decreasing:

$$\begin{array}{c}\frac{d}{dx}{a}_x=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}\\ \\ \gg \gg \frac{2\left(\sqrt{x}-\sqrt{x+1}\right)}{4\sqrt{x}\sqrt{x+1}}\end{array}$$

Denominator is $>0$. Numerator is $<0$. So $\frac{d}{dx}{a}_x<0$ and $a_x$ is monotone decreasing.

Therefore $a_n$ is monotone decreasing as $n\to \infty$.

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(3) Since $a_n>0$ it is bounded below by $0$. It is monotone decreasing. Conclude that it converges.

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Theory

Theory 3

Series convergence

We say that a series converges when its partial sum sequence converges:

$$\text{“}\sum _{n=0}^{\infty }{a}_n\text{converges}\text{”}\text{MEANS:}\text{“}{S}_N\text{converges as N→∞”}$$

Let us apply this to the geometric series. Recall our formula for the partial sums:

$$S_N=a_0\frac{1-r^{N+1}}{1-r}$$

Rewrite this formula:

$$\gg \gg S_N=\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{N+1}$$

Now take the limit as $N\to \infty$:

$$\underset{N\to \infty }{\lim }{S}_N=\text{“}\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{\infty +1}\text{”}=\frac{a_0}{1-r}$$

So we see that $S_N$ converges exactly when $|r|<1$. It converges to $\frac{a_0}{1-r}$.

(If $|r|=1$ then the denominator is $0$, and if $|r|>1$ then the factor $r^{\infty +1}$ does not converge.)

Furthermore, we have the limit value:

$$\sum _{n=0}^{\infty }{a}_n\gg \gg \underset{N\to \infty }{\lim }{S}_N\gg \gg \frac{a_0}{1-r}\gg \gg S$$

This result confirms the formula we derived for the total sum $S$ of a geometric series. This time we did not start by assuming $S$ exists, rather we proved that $S$ exists. (Provided that $|r|<1$.)

Extra - Looking at $S$ and $S_N$

Notice that we always have the rule:

$$\begin{array}{rl}{S}_N& =S-r^{N+1}S\\ & \\ S_N& =\frac{a_0}{1-r}-\frac{a_0}{1-r}{r}^{N+1}\end{array}$$

This rule can be viewed as coming from partitioning the full series into a finite part $S_N$ and the remaining infinite part:

$$\begin{array}{rl}S& =\underset{S_N}{\underbrace{a_0+a_{0}r+\cdots +a_0{r}^N}}+\underset{S-S_N}{\underbrace{a_0{r}^{N+1}+a_0{r}^{N+2}+\dots }}\end{array}$$

We can remove a factor $r^{N+1}$ from the infinite part:

$$S-S_N=r^{N+1}\left(a_0+a_{0}r+a_0{r}^2\dots \right)$$

The parenthetical expression is equal to $S$, so we have the formula $S_N=S-r^{N+1}S$ given above.

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Simple divergence test

Videos

Review Videos

Videos, Math Dr. Bob

• Geometric series and SDT, again: Geometric series, Simple Divergence Test (aka “Limit Test”)
• Integral test: Basics
• Integral test: $p$-series
  • Extra: Integral test: Further examples
  • Extra: Integral test: Estimations

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Theory

Theory 1

Simple Divergence Test (SDT)

Applicability: Any series.

Test Statement:

$$\underset{n\to \infty }{\lim }{a}_n\ne 0⟹\sum _{n=1}^{\infty }{a}_n\text{diverges}$$

AKA the “Not Even Close” test

The converse is not valid. For example, ${\sum }_{n=1}^{\infty }\frac{1}{n}$ diverges even though ${\lim }_{n\to \infty }\frac{1}{n}=0$.

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Illustration

Example - Simple Divergence Test: examples

Simple divergence test: examples

(a) Consider: $\sum _{n=1}^{\infty }\frac{n}{4n+1}$

This diverges by the SDT because $a_n\to \frac{1}{4}$ and not $0$.

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(b) Consider: $\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{n}{n+1}$

This diverges by the SDT because ${\lim }_{n\to \infty }{a}_n=\text{DNE}$.

We can say the terms “converge to the pattern $+1,-1,+1,-1,\dots$,” but that is not a limit value.

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Positive series

Videos

Review Videos

• Direct Comparison Test: Theory and basic examples
• Direct Comparison Test: Series $\frac{1}{\ln n}$
• Limit Comparison Test: Theory and basic examples
• Limit Comparison Test: Further examples

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Theory

Theory 1

Positive series

A series is called positive when its individual terms are positive, i.e. $a_n>0$ for all $n$.

The partial sum sequence $S_N$ is monotone increasing for a positive series.

By the monotonicity test for convergence of sequences, $S_N$ therefore converges whenever it is bounded above. If $S_N$ is not bounded above, then ${\sum }_{n=1}^{\infty }{a}_n$ diverges to $+\infty$.

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Another test, called the integral test, studies the terms of a series as if they represent rectangles with upper corner pinned to the graph of a continuous function.

To apply the test, we must convert the integer index variable $n$ in $a_n$ into a continuous variable $x$. This is easy when we have a formula for $a_n$ (provided it doesn’t contain factorials or other elements dependent on integrality).

Integral Test (IT)

Applicability: $f(x)$ must be:

• Continuous
• Positive
• Monotone decreasing

Test Statement:

$$\sum _{n=1}^{\infty }{a}_n\text{converges}⟺{\int }_1^{\infty }f(x)dx\text{converges}$$

Extra - Integral test: explanation

To show that integral convergence implies series convergence, consider the diagram:

This shows that ${\sum }_{n=2}^N{a}_n\le {\int }_1^{N}f(x)dx$ for any $N$. Therefore, if ${\int }_1^{\infty }f(x)dx$ converges, then ${\int }_1^{N}f(x)dx$ is bounded (independent of $N$) and so ${\sum }_{n=2}^N{a}_n$ is bounded by that inequality. But ${\sum }_{n=2}^N{a}_n=S_N-a_1$; so by adding $a_1$ to the bound, we see that $S_N$ itself is bounded, which implies that ${\sum }_{n=1}^{\infty }{a}_n$ converges.

To show that integral divergence implies series divergence, consider a similar diagram:

This shows that ${\sum }_{n=1}^{N-1}{a}_n\ge {\int }_1^{N}f(x)dx$ for any $N$. Therefore, if ${\int }_1^{\infty }f(x)dx$ diverges, then ${\int }_1^{N}f(x)dx$ goes to $+\infty$ as $N\to \infty$, and so ${\sum }_{n=1}^{N-1}{a}_n$ goes to $+\infty$ as well. So ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

Notice: the picture shows $f(x)$ entirely above (or below) the rectangles. This depends upon $f(x)$ being monotone decreasing, as well as $f(x)>0$. (This explains the applicability conditions.)

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Next we use the integral test to evaluate the family of $p$-series, and later we can use $p$-series in comparison tests without repeating the work of the integral test.

$p$-series

A $p$-series is a series of this form: $\sum _{n=1}^{\infty }\frac{1}{n^p}$

Convergence properties:

$$p>1:\text{series converges}p\le 1:\text{series diverges}$$

Extra - Proof of $p$-series convergence

(1) To verify the convergence properties of $p$-series, apply the integral test:

Applicability: verify it’s continuous, positive, decreasing.

Convert $n$ to $x$ to obtain the function $f(x)=\frac{1}{x^p}$.

Indeed $\frac{1}{x^p}$ is continuous and positive and decreasing as $x$ increases.

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(2) Apply the integral test.

Integrate, assuming $p\ne 1$:

$$\begin{array}{rl}{\int }_1^{\infty }\frac{1}{x^p}dx& \gg \gg \underset{R\to \infty }{\lim }{\frac{x^{p-1}}{p-1}|}_1^R\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\left(\frac{R^{-p+1}}{-p+1}-\frac{1^{-p+1}}{-p+1}\right)\end{array}$$

When $p>1$ we have ${\lim }_{R\to \infty }\frac{R^{-p+1}}{-p+1}=0$

When $p<1$ we have ${\lim }_{R\to \infty }\frac{R^{-p+1}}{-p+1}=\infty$

When $p=1$, integrate a second time:

$$\begin{array}{rl}{\int }_1^{\infty }\frac{1}{x}dx& \gg \gg \underset{R\to \infty }{\lim }\ln x{|}_1^R\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\ln R-\ln 1\gg \gg \infty \end{array}$$

Conclude: the integral converges when $p>1$ and diverges when $p\le 1$.

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We could instead immediately refer to the convergence results for $p$-integrals instead of reproving them here.

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Illustration

Example - $p$-series examples

p-series examples

By finding $p$ and applying the $p$-series convergence properties:

We see that ${\sum }_{n=1}^{\infty }\frac{1}{n^{1.1}}$ converges: $p=1.1$ so $p>1$

But ${\sum }_{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges: $p=1/2$ so $p\le 1$

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Example - Integral test - pushing the envelope of convergence

Integral test - pushing the envelope of convergence

Does $\sum _{n=2}^{\infty }\frac{1}{n\ln n}$ converge?

Does $\sum _{n=2}^{\infty }\frac{1}{n(\ln n{)}^2}$ converge?

Notice that $\ln n$ grows very slowly with $n$, so $\frac{1}{n\ln n}$ is just a little smaller than $\frac{1}{n}$ for large $n$, and similarly $\frac{1}{n(\ln n{)}^2}$ is just a little smaller still.

Solution

(1) The two series lead to the two functions $f(x)=\frac{1}{x\ln x}$ and $g(x)=\frac{1}{x(\ln x{)}^2}$.

Check applicability.

Clearly $f(x)$ and $g(x)$ are both continuous, positive, decreasing functions on $x\in [2,\infty )$.

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(2) Apply the integral test to $f(x)$.

Integrate $f(x)$:

$$\begin{array}{rl}{\int }_2^{\infty }\frac{1}{x\ln x}dx& \gg \gg {\int }_{u=\ln 2}^{\infty }\frac{1}{u}du\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\ln u{|}_{\ln 1}^R\gg \gg \infty \end{array}$$

Conclude: ${\sum }_{n=2}^{\infty }\frac{1}{n\ln n}$ diverges.

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(3) Apply the integral test to $g(x)$.

Integrate $g(x)$:

$$\begin{array}{rl}{\int }_2^{\infty }\frac{1}{x(\ln x{)}^2}dx& \gg \gg {\int }_{u=\ln 2}^{\infty }\frac{1}{u^2}du\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }-u^{-1}{|}_{\ln 2}^R\gg \gg \frac{1}{\ln 2}\end{array}$$

Conclude: ${\sum }_{n=2}^{\infty }\frac{1}{n(\ln n{)}^2}$ converges.

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Theory

Theory 2

Direct Comparison Test (DCT)

Applicability: Both series are positive: $a_n>0$ and $b_n>0$.

Test Statement: Suppose $a_n\le b_n$ for large enough $n$.
(Meaning: for $n\ge N$ with some given $N$.) Then:

• Smaller pushes up bigger: $$\sum _{n=1}^{\infty }{a}_n\text{diverges}⟹\sum _{n=1}^{\infty }{b}_n\text{diverges}$$
• Bigger controls smaller: $$\sum _{n=1}^{\infty }{b}_n\text{converges}⟹\sum _{n=1}^{\infty }{a}_n\text{converges}$$

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Illustration

Example - Direct comparison test: rational functions

Direct comparison test: rational functions

(a) $\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}{3}^n}$

Choose: $a_n=\frac{1}{\sqrt{n}{3}^n}$ and $b_n=\frac{1}{3^n}$

Check: $\frac{1}{\sqrt{n}{3}^n}\le \frac{1}{3^n}$

Observe: $\sum \frac{1}{3^n}$ is a convergent geometric series

Therefore: converges by the DCT.

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(b) $\sum _{n=1}^{\infty }\frac{{\cos }^{2}n}{n^3}$

Choose: $a_n=\frac{{\cos }^{2}n}{n^3}$ and $b_n=\frac{1}{n^3}$.

Check: $\frac{{\cos }^{2}n}{n^3}\le \frac{1}{n^3}$

Observe: $\sum \frac{1}{n^3}$ is a convergent $p$-series

Therefore: converges by the DCT.

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(c) $\sum _{n=1}^{\infty }\frac{n}{n^3+1}$

Choose: $a_n=\frac{n}{n^3+1}$ and $b_n=\frac{1}{n^2}$

Check: $\frac{n}{n^3+1}\le \frac{1}{n^2}$ (notice that $\frac{n}{n^3+1}\le \frac{n}{n^3}$)

Observe: $\sum \frac{1}{n^2}$ is a convergent $p$-series

Therefore: converges by the DCT.

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(d) $\sum _{n=2}^{\infty }\frac{1}{n-1}$

Choose: $a_n=\frac{1}{n}$ and $b_n=\frac{1}{n-1}$

Check: $\frac{1}{n}\le \frac{1}{n-1}$

Observe: $\sum \frac{1}{n}$ is a divergent $p$-series

Therefore: diverges by the DCT.

Link to original

Theory

Theory 3

Some series can be compared using the DCT after applying certain manipulations and tricks.

For example, consider the series ${\sum }_{n=2}^{\infty }\frac{1}{n^2-1}$. We suspect convergence because $a_n\approx \frac{1}{n^2}$ for large $n$. But unfortunately, $a_n>\frac{1}{n^2}$ always, so we cannot apply the DCT.

We could make some ad hoc arguments that do use the DCT, eventually:

Trick Method 1:

• Observe that for $n>1$ we have $\frac{1}{n^2-1}\le \frac{10}{n^2}$. (Check it!)
• But $\sum \frac{10}{n^2}$ converges, indeed its value is $10\cdot \sum \frac{1}{n^2}$, which is $\frac{10{\pi }^2}{6}$.
• So the series $\sum \frac{1}{n^2-1}$ converges.

Trick Method 2:

• Observe that we can change the letter $n$ to $n+1$ by starting the new $n$ at $n=1$.
• Then we have: $$\sum _{n=2}^{\infty }\frac{1}{n^2-1}=\sum _{n=1}^{\infty }\frac{1}{(n+1{)}^2-1}=\sum _{n=1}^{\infty }\frac{1}{n^2+2n}$$
• This last series has terms smaller than $\frac{1}{n^2}$ so the DCT with $b_n=\frac{1}{n^2}$ (a convergent $p$-series) shows that the original series converges too.

These convoluted arguments suggest that a more general version of Comparison is possible.

Indeed, it is sufficient to compare the relative large-n behavior of the two series. We use the termwise ratios to estimate comparative behavior for increasing $n$.

Limit Comparison Test (LCT)

Applicability: Both series are positive: $a_n>0$ and $b_n>0$.

Test Statement: Suppose that ${\lim }_{n\to \infty }\frac{a_n}{b_n}=L$. Then:
If $0<L<\infty$, i.e. finite non-zero, then:

$$\sum a_n\text{converges}⟺\sum b_n\text{converges}$$

Extra - LCT edge cases

If $L=0$ or $L=\infty$, we can still draw an inference, but only in one direction:

• If $L=0$: $$\sum b_n\text{converges}⟹\sum a_n\text{converges}$$
• If $L=\infty$: $$\sum b_n\text{diverges}⟹\sum a_n\text{diverges}$$

Extra - Limit Comparison Test explanation

Suppose $a_n/b_n\to L$ and $0<L<\infty$. Then for $n$ sufficiently large, we know $a_n/b_n<L+1$.

Doing some algebra, we get $a_n<(L+1)b_n$ for $n$ large.

If $\sum b_n$ converges, then $\sum (L+1)b_n$ also converges (constant multiple), and then the DCT implies that $\sum a_n$ converges.

Conversely: we also know that $b_n/a_n\to 1/L$, so $b_n<(1/L+1)a_n$ for all $n$ sufficiently large. Thus if $\sum a_n$ converges, $\sum (1/L+1)a_n$ also converges, and by the DCT again $\sum b_n$ converges too.

The cases with $L=0$ or $L=\infty$ are handled similarly.

Link to original

Illustration

Example - Limit Comparison Test examples

Limit comparison test examples

(a) $\sum _{n=1}^{\infty }\frac{1}{2^n-1}$

Choose: $a_n=\frac{1}{2^n-1}$ and $b_n=\frac{1}{2^n}$.

Compare in the limit:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}\gg \gg \underset{n\to \infty }{\lim }\frac{2^n}{2^n-1}\gg \gg 1=:L$$

Observe: $\sum \frac{1}{2^n}$ is a convergent geometric series

Therefore: converges by the LCT.

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(b) $\sum _{n=1}^{\infty }\frac{2n^2+3n}{\sqrt{5+n^5}}$

Choose: $a_n=\frac{2n^2+3n}{\sqrt{5+n^5}}$, $b_n=\frac{1}{n^{1/2}}$

Compare in the limit:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}\gg \gg \underset{n\to \infty }{\lim }\frac{(2n^2+3n)\sqrt{n}}{\sqrt{5+n^5}}$$ $$\frac{(2n^2+3n)\sqrt{n}}{\sqrt{5+n^5}}\stackrel{n\to \infty }{⟶}\frac{2n^{5/2}}{n^{5/2}}\to 2=:L$$

Observe: $\sum \frac{1}{n^{1/2}}$ is a divergent $p$-series

Therefore: diverges by the LCT.

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(c) $\sum _{n=2}^{\infty }\frac{n^2}{n^4-n-1}$

Choose: $a_n=\frac{n^2}{n^4-n-1}$ and $b_n=\frac{1}{n^2}$

Compare in the limit:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}\gg \gg \underset{n\to \infty }{\lim }\frac{n^4}{n^4-n-1}\gg \gg 1=:L$$

Observe: ${\sum }_{n=2}^{\infty }\frac{1}{n^2}$ is a converging $p$-series

Therefore: converges by the LCT.

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Alternating series

Videos

Review Videos

Videos, Math Dr. Bob:

• Alternating Series Test: Theory and basic examples
• Alternating Series Test: Remainder estimates
• Alternating Series Test: Further remainder estimates

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Theory

Theory 1

Consider these series:

$$\begin{array}{rl}1& +\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots =\infty \\ & \\ -1& -\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\cdots =-\infty \\ & \\ 1& -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\cdots =\ln 2\\ & \\ 1& +\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\cdots =?\end{array}$$

The absolute values of terms are the same between these series, only the signs of terms change.

The first is a positive series because there are no negative terms.

The second series is the negation of a positive series – the study of such series is equivalent to that of positive series, just add a negative sign everywhere. These signs can be factored out of the series. (For example $\sum -\frac{1}{n}=-\sum \frac{1}{n}$.)

The third series is an alternating series because the signs alternate in a strict pattern, every other sign being negative.

The fourth series is not alternating, nor is it positive, nor negative: it has a mysterious or unknown pattern of signs.

A series with any negative signs present, call it ${\sum }_{n=1}^{\infty }{a}_n$, converges absolutely when the positive series of absolute values of terms, namely ${\sum }_{n=1}^{\infty }|a_n|$, converges.

THEOREM: Absolute implies ordinary

If a series ${\sum }_{n=1}^{\infty }{a}_n$ converges absolutely, then it also converges as it stands.

A series might converge due to the presence of negative terms and yet not converge absolutely:

A series ${\sum }_{n=1}^{\infty }{a}_n$ is said to be converge conditionally when the series converges as it stands, but the series produced by inserting absolute values, namely ${\sum }_{n=1}^{\infty }|a_n|$, diverges.

The alternating harmonic series above, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2$, is therefore conditionally convergent. Let us see why it converges. We can group the terms to create new sequences of pairs, each pair being a positive term. This can be done in two ways. The first creates an increasing sequence, the second a decreasing sequence:

$$\begin{array}{rl}\text{increasing from 0:}& \left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\cdots \\ & \\ \text{decreasing from 1:}& 1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\cdots \end{array}$$

Suppose $S_N$ gives the sequence of partial sums of the original series. Then $S_{2N}$ gives the first sequence of pairs, namely $S_2$, $S_4$, $S_6$, $\dots$ . And $S_{2N-1}$ gives the second sequence of pairs, namely $S_1$, $S_3$, $S_5$, $\dots$ .

The second sequence shows that $S_N$ is bounded above by $1$, so $S_{2N}$ is monotone increasing and bounded above, so it converges. Similarly $S_{2N-1}$ is monotone decreasing and bounded below, so it converges too, and of course they must converge to the same thing.

The fact that the terms were decreasing in magnitude was an essential ingredient of the argument for convergence. This fact ensured that the parenthetical pairs were positive numbers.

Alternating Series Test (AST) - “Leibniz Test”

Applicability: Alternating series only: ${\sum }_{n=1}^{\infty }(-1{)}^{n-1}{a}_n$ with $a_n>0$

Test Statement:
If:

1. $a_n\to 0$ as $n\to \infty$ (i.e. it passes the SDT: if this fails, conclude diverges)
2. $a_n$ are decreasing, so $a_1>a_2>a_3>a_4>\cdots >0$

Then:

$$\sum _{n=1}^{\infty }(-1{)}^{n-1}{a}_n\text{converges}$$

“Next Term Bound” rule for error of the partial sums:

$$|S-S_N|<a_{N+1}$$

Extra - Alternating Series Test: Theory

Just as for the alternating harmonic series, we can form positive paired-up series because the terms are decreasing:

$$\begin{array}{r}\left(a_1-a_2\right)+\left(a_3-a_4\right)+\left(a_5-a_6\right)+\cdots \\ \\ a_1-\left(a_2-a_3\right)-\left(a_4-a_5\right)-\left(a_6-a_7\right)-\cdots \end{array}$$

The first sequence $S_{2N}$ is monotone increasing from $0$, and the second $S_{2N-1}$ is decreasing from $a_1$. The first is therefore also bounded above by $a_1$. So it converges. Similarly, the second converges. Their difference at any point is $S_{2N}-S_{2N-1}$ which is equal to $-a_{2N}$, and this goes to zero. So the two sequences must converge to the same thing.

By considering these paired-up sequences and the effect of adding each new term one after the other, we obtain the following order relations:

$$0<S_2<S_4<S_6<\cdots <S<\cdots <S_5<S_3<S_1=a_1$$

Thus, for any even $2N$ and any odd $2M-1$:

$$S_{2N}<S<S_{2M-1}$$

Now set $M=N$ and subtract $S_{2N-1}$ from both sides:

$$\begin{array}{rl}{S}_{2N}-S_{2N-1}& <S-S_{2N-1}<0\\ & \\ \gg \gg -a_{2N}& <S-S_{2N-1}<0\end{array}$$

Now set $M=N+1$ and subtract $S_{2N}$ from both sides:

$$\begin{array}{rl}0<S-S_{2N}& <S_{2N+1}-S_{2N}\\ & \\ \gg \gg 0<S-S_{2N}& <a_{2N+1}\end{array}$$

This covers both even cases ($n=2N$) and odd cases ($n=2N-1$). In either case, we have:

$$|S-S_n|<a_{n+1}$$

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Illustration

Example - Alternating Series Test: Basic illustration

Alternating series test: basic illustration

(a) $\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}}{\sqrt{n}}$ converges by the AST.

Notice that $\sum \frac{1}{\sqrt{n}}$ diverges as a $p$-series with $p=1/2<1$.

Therefore the first series converges conditionally.

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(b) $\sum _{n=1}^{\infty }\frac{\cos n\pi }{n^2}$ converges by the AST.

Notice the funny notation: $\cos n\pi =(-1{)}^n$.

This series converges absolutely because $\left|\frac{\cos n\pi }{n^2}\right|=\frac{1}{n^2}$, which is a $p$-series with $p=2>1$.

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Example - Approximating $\pi$

Approximating pi

The Taylor series for ${\tan }^{-1}x$ is given by:

$${\tan }^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

Using the AST “next term bound,” how many terms are needed to approximate $\pi$ to within $0.001$?

Solution

(1) The main idea is to use $\tan \frac{\pi }{4}=1$ and thus ${\tan }^{-1}1=\frac{\pi }{4}$. Therefore:

$$\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$

and thus:

$$\pi =4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\cdots$$

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(2) Write $E_n$ for the error of the approximation, meaning $E_n=S-S_n$.

By the AST error formula, we have $|E_n|<a_{n+1}$.

We desire $n$ such that $|E_n|<0.001$. Therefore, calculate $n$ such that $a_{n+1}<0.001$, and then we will know:

$$|E_n|<a_{n+1}<0.001$$

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(3) The general term is $a_n=\frac{4}{2n-1}$. Plug in $n+1$ in place of $n$ to find $a_{n+1}=\frac{4}{2n+1}$. Now solve:

$$\begin{array}{rl}& a_{n+1}=\frac{4}{2n+1}<0.001\\ & \\ \gg \gg & \frac{4}{0.001}<2n+1\\ & \\ \gg \gg & 3999<2n\\ & \\ \gg \gg & 2000\le n\end{array}$$

We conclude that at least $2000$ terms are necessary to be confident (by the error formula) that the approximation of $\pi$ is accurate to within $0.001$.

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Ratio test and Root test

Videos

Review Videos

Videos, Math Dr. Bob

• Ratio test: Basics
• Ratio test: Ratio test + DCT
• Root test: Basics
• Root test: for $\sum (1-1/n^2{)}^{n^3}$

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Theory

Theory 1

Ratio Test (RaT)

Applicability: Any series with nonzero terms.

Test Statement:

Suppose that $\left|\frac{a_{n+1}}{a_n}\right|⟶L$ as $n\to \infty$.

Then:

$$\begin{array}{rlrl}L<1:& & & \sum _{n=1}^{\infty }{a}_n\text{converges absolutely}\\ L>1:& & & \sum _{n=1}^{\infty }{a}_n\text{diverges}\\ & & & \\ L=1\text{or DNE}:& & & \text{test inconclusive}\end{array}$$

Extra - Ratio Test - a deeper look

To understand the ratio test better, first consider this series:

$$\sum _{n=0}^{\infty }\frac{2^n}{n!}=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\cdots$$

• The term $\frac{2^3}{3!}$ is created by multiplying the prior term by $\frac{2}{3}$.
• The term $\frac{2^4}{4!}$ is created by multiplying the prior term by $\frac{2}{4}$.
• The term $a_n$ is created by multiplying the prior term by $\frac{2}{n}$.

When $n>3$, the multiplication factor giving the next term is necessarily less than $\frac{2}{3}$. Therefore, when $n>3$, the terms shrink faster than those of a geometric series having $r=\frac{2}{3}$. Therefore this series converges.

Similarly, consider this series:

$$\sum _{n=0}^{\infty }\frac{10^n}{n!}=1+\frac{10}{1!}+\frac{10^2}{2!}+\frac{10^3}{3!}+\cdots$$

Write $R_n=\frac{a_n}{a_{n-1}}$ for the ratio from the prior term $a_{n-1}$ to the current term $a_n$. For this series, $R_n=\frac{10}{n}$.

This ratio falls below $\frac{10}{11}$ when $n>11$, after which the terms necessarily shrink faster than those of a geometric series with $r=\frac{10}{11}$. Therefore this series converges.

The main point of the discussion can be stated like this:

$$R_n\to L<1\text{as}n\to \infty$$

Whenever this is the case, then eventually the ratios are bounded below some $r<1$, and the series terms are smaller than those of a converging geometric series.

Extra - Ratio Test proof

Let us write $R_n=\left|\frac{a_{n+1}}{a_n}\right|$ for the ratio to the next term from term $n$.

Suppose that $R_n\to L$ as $n\to \infty$, and that $L<1$. This means: eventually the ratio of terms is close to $L$; so eventually it is less than $1$.

More specifically, let us define $r=\frac{L+1}{2}$. This is the point halfway between $L$ and $1$. Since $R_n\to L$, we know that eventually $R_n<r$.

Any geometric series with ratio $r$ converges. Set $c=a_N$ for $N$ big enough that $R_N<r$. Then the terms of our series satisfy $|a_{N+n}|\le c{r}^n$, and the series starting from $a_N$ is absolutely convergent by comparison to this geometric series.

(Note that the terms $a_1,\dots ,a_{N-1}$ do not affect convergence.)

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Illustration

Example - Ratio test

Ratio test examples

(a) Observe that $\sum _{n=0}^{\infty }\frac{10^n}{n!}$ has ratio $R_n=\frac{10}{n+1}$ and thus $R_n\to 0=L<1$. Therefore the RaT implies that this series converges.

Simplify the ratio:

$$\begin{array}{c}\frac{\left(\frac{10^{n+1}}{(n+1)!}\right)}{\left(\frac{10^n}{n!}\right)}\gg \gg \frac{10^{n+1}}{(n+1)!}\cdot \frac{n!}{10^n}\\ \\ \gg \gg \frac{10\cdot 10^n}{(n+1)n!}\cdot \frac{n!}{10^n}\gg \gg \frac{10}{n+1}\stackrel{n\to \infty }{⟶}0\end{array}$$

Notice this technique! We frequently use these rules:

$$10^{n+1}=10^n\cdot 10,(n+1)!=(n+1)n!$$

(To simplify ratios with exponents and factorials.)

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(b) $\sum _{n=1}^{\infty }\frac{n^2}{2^n}$ has ratio $R_n=\frac{(n+1{)}^2}{2^{n+1}}/\frac{n^2}{2^n}$.

Simplify this:

$$\frac{(n+1{)}^2}{2^{n+1}}/\frac{n^2}{2^n}\gg \gg \frac{(n+1{)}^2}{2^{n+1}}\cdot \frac{2^n}{n^2}$$ $$\gg \gg \frac{(n+1{)}^2\cdot 2^n}{n^2\cdot 2\cdot 2^n}\gg \gg \frac{n^2+2n+1}{2n^2}\stackrel{n\to \infty }{⟶}\frac{1}{2}=L$$

So the series converges absolutely by the ratio test.

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(c) Observe that $\sum _{n=1}^{\infty }{n}^2$ has ratio $R_n=\frac{n^2+2n+1}{n^2}\to 1$ as $n\to \infty$.

So the ratio test is inconclusive, even though this series fails the SDT and obviously diverges.

---

(d) Observe that $\sum _{n=1}^{\infty }\frac{1}{n^2}$ has ratio $R_n=\frac{n^2}{n^2+2n+1}\to 1$ as $n\to \infty$.

So the ratio test is inconclusive, even though the series converges as a $p$-series with $p=2>1$.

---

(e) More generally, the ratio test is usually inconclusive for rational functions; it is more effective to use LCT with a $p$-series.

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Theory

Theory 2

Root Test (RooT)

Applicability: Any series.

Test Statement:

Suppose that ${|a_n|}^{1/n}⟶L$ as $n\to \infty$.

Then:

$$\begin{array}{rlrl}L<1:& & & \sum _{n=1}^{\infty }{a}_n\text{converges absolutely}\\ L>1:& & & \sum _{n=1}^{\infty }{a}_n\text{diverges}\\ & & & \\ L=1\text{or DNE}:& & & \text{test inconclusive}\end{array}$$

Extra - Root test: explanation

The fact that ${|a_n|}^{1/n}\to L$ and $L<1$ implies that eventually ${|a_n|}^{1/n}<\frac{L+1}{2}$ for all high enough $n$. Set $r=\frac{L+1}{2}$ (the midpoint between $L$ and $1$).

Now, the equation ${|a_n|}^{1/n}<r$ is equivalent to the equation $|a_n|<r^n$.

Therefore, eventually the terms $|a_n|$ are each less than the corresponding terms of this convergent geometric series:

$$\sum _{n=1}^{\infty }{r}^n=1+r+r^2+r^3+\cdots$$

Link to original

Illustration

Example - Root test examples

Root test examples

(a) Observe that $\sum _{n=1}^{\infty }{\left(\frac{1}{n}\right)}^n$ has roots of terms:

$$|a_n{|}^{1/n}={\left({\left(\frac{1}{n}\right)}^n\right)}^{1/n}=\frac{1}{n}\stackrel{n\to \infty }{⟶}0=L$$

Because $L<1$, the RooT shows that the series converges absolutely.

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(b) Observe that $\sum _{n=1}^{\infty }(-1{)}^n{\left(\frac{n}{2n+1}\right)}^n$ has roots of terms:

$$\sqrt[n]{|a_n|}=\frac{n}{2n+1}\stackrel{n\to \infty }{⟶}\frac{1}{2}=L$$

Because $L<1$, the RooT shows that the series converges absolutely.

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Example - Ratio test versus root test

Ratio test versus root test

Determine whether the series $\sum _{n=1}^{\infty }\frac{n^2{4}^n}{5^{n+2}}$ converges absolutely or conditionally or diverges.

Solution
Before proceeding, rewrite somewhat the general term as ${\left(\frac{n}{5}\right)}^2\cdot {\left(\frac{4}{5}\right)}^n$.

Now we solve the problem first using the ratio test. By plugging in $n+1$ we see that

$$a_{n+1}={\left(\frac{n+1}{5}\right)}^2\cdot {\left(\frac{4}{5}\right)}^{n+1}$$

So for the ratio $R_n$ we have:

$$\begin{array}{c}{a}_{n+1}\cdot \frac{1}{a_n}={\left(\frac{n+1}{5}\right)}^2{\left(\frac{4}{5}\right)}^{n+1}\cdot {\left(\frac{5}{n}\right)}^2{\left(\frac{5}{4}\right)}^n\\ \\ \gg \gg \frac{n^2+2n+1}{n^2}\cdot \frac{4}{5}⟶\frac{4}{5}<1\text{ as }n\to \infty \end{array}$$

Therefore the series converges absolutely by the ratio test.

Now solve the problem again using the root test. We have for $\sqrt[n]{|a_n|}$:

$${\left({\left(\frac{n}{5}\right)}^2{\left(\frac{4}{5}\right)}^n\right)}^{1/n}={\left(\frac{n}{5}\right)}^{2/n}\frac{4}{5}$$

To compute the limit as $n\to \infty$ we must use logarithmic limits and L’Hopital’s Rule. So, first take the log:

$$\ln \left({\left(\frac{n}{5}\right)}^{2/n}\frac{4}{5}\right)=\frac{2}{n}\ln \frac{n}{5}+\ln \frac{4}{5}$$

Then for the first term apply L’Hopital’s Rule:

$$\frac{\ln \frac{n}{5}\stackrel{d/dx}{⟶}\frac{1}{n/5}\frac{1}{5}}{n/2\stackrel{d/dx}{⟶}1/2}\gg \gg \frac{1/n}{1/2}\gg \gg \frac{2}{n}⟶0\text{ as }n\to \infty$$

So the first term goes to zero, and the second (constant) term is the value of the limit. So the log limit is $\ln \frac{4}{5}$, and the limit (before taking logs) must be $e^{\ln \frac{4}{5}}$ (inverting the log using $e^x$) and this is $\frac{4}{5}$. Since $\frac{4}{5}<1$, the root test also shows that the series converges absolutely.

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Series tests: strategy tips

Videos

Review Videos

Videos, Math Dr. Bob

• Series test round-up: Part I
• Series test round-up: Part II
• Series test round-up: Part III

Videos, Trefor Bazett

• How to choose a series convergence test

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Theory

Theory 1

It can help to associate certain “strategy tips” to find convergence tests based on certain patterns.

Matching powers → Simple Divergence Test

$$\sum _{n=1}^{\infty }\frac{n-1}{2n+1}$$

Use the SDT because we see the highest power is the same ($=1$) in numerator and denominator.

Rational or Algebraic → Limit Comparison Test

$$\sum _{n=1}^{\infty }\frac{\sqrt{n^3+1}}{3n^3+4n^2+2}$$

Use the LCT because we have a rational or algebraic function (positive terms).

Not rational, not factorials → Integral Test

$$\sum _{n=1}^{\infty }n{e}^{-n^2}$$

Use the IT because we do not have a rational/algebraic function, and we do not see factorials.

Rational, alternating → AST, and LCT or DCT

$$\sum _{n=1}^{\infty }(-1{)}^n\frac{n^2}{n^4+1}$$

Notice large $n$ behavior is like $n^2/n^4$ or $1/n^2$. This converges. Use the LCT to show absolute convergence. Skip the AST because absolute convergence settles the matter. Lesson: check for absolute BEFORE applying AST, even when alternating!

Factorials → Ratio Test

$$\sum _{n=1}^{\infty }\frac{2^n}{n!}$$

Use the RaT because we see a factorial. (In case of alternating + factorial, use RaT first.)

Recognize geometric → LCT or DCT

$$\sum _{n=1}^{\infty }\frac{1}{2+3^n}$$

Use the LCT or DCT comparing to $\frac{1}{3^n}$ because we see similarity to $\frac{1}{3^n}$ (recognize geometric).

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