W09 - Examples

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.

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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 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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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 examples

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.

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(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 $p=2>1$.

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(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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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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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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