W07 - Examples

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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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

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

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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Monotonicity

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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