Unit 03 - Essential problems

Sequences

01

L’Hopital practice - converting indeterminate form

By imitating the technique of the L’Hopital’s Rule example, find the limit of the sequence:

$$a_n=\sqrt{n}\ln (1+\frac{1}{n})$$

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Solution

Solutions - 0140-01

(1) Indeterminate form:

$$a_n\to \infty \cdot 0\text{as}n\to \infty$$

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(2) L’Hopital:

Convert:

$$a_n=\sqrt{n}\ln (1+\frac{1}{n})\gg \gg \frac{\ln (1+1/n)}{n^{-1/2}}\to \frac{0}{0}$$

Change to $x$ and apply L’Hopital:

$$\frac{\ln (1+1/x)}{x^{-1/2}}\stackrel{d/dx}{⟶}\frac{\frac{-1/x^2}{1+1/x}}{-1/2\cdot x^{-3/2}}\gg \gg \frac{2}{x^{+1/2}+x^{-1/2}}$$

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(3) Take limit:

$$\frac{2}{x^{-1/2}+x^{+1/2}}⟶0\text{as}x\to \infty$$

Therefore $a_n\to 0$ as $n\to \infty$.

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06

Limits and convergence

For each sequence, either write the limit value (if it converges), or write ‘diverges’.

(a) ${1.01}^n$ $$ (b) ${\displaystyle 2^{1/n}}$ $$ (c) ${\displaystyle \frac{n!}{9^n}}$ $$ (d) ${\displaystyle \frac{3n^2+n+2}{2n^2-3}}$

(e) ${\displaystyle \frac{\cos n}{n}}$ $$ (f) $\ln 5^n-\ln n!$ $$ (g) ${\displaystyle {(2+\frac{4}{n^2})}^{1/3}}$ $$ (h) ${\displaystyle n\sin \frac{\pi }{n}}$

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Solution

Solutions - 0140-06

(a) diverges $$ (b) $1$ $$ (c) diverges $$ (d) ${\displaystyle \frac{3}{2}}$

(e) $0$

(f) diverges:

$$\begin{array}{c}\ln 5^n-\ln n!\\ \\ \gg \gg \ln \left(\frac{5^n}{n!}\right)\stackrel{n\to \infty }{⟶}\ln (0^{+})=-\infty \end{array}$$

(g) $2^{1/3}$

(h) $0$

Use L’Hopital’s Rule:

$$\begin{array}{c}n\sin \frac{\pi }{n}\gg \gg \frac{\sin \frac{\pi }{n}}{1/n}\gg \gg \frac{\sin \frac{\pi }{x}}{1/x}\\ \\ {\underset{d/dx}{⟶}}^{d/dx}\frac{(-\pi /x^2)\cdot \cos \frac{\pi }{x}}{-1/x^2}\gg \gg \pi \cos \frac{\pi }{x}\stackrel{x\to \infty }{⟶}\pi \end{array}$$Link to original

Series basics

02

Geometric series

Compute the following summation values using the sum formula for geometric series.

(a) ${\displaystyle \sum _{n=0}^{\infty }{5}^{-n}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{2+3^n}{5^n}}$

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Solution

Solutions - 0150-02

(a)

The first term is $a_0=5^{-0}=1$. The common ratio is $1/5$. Therefore the sum:

$$S=\frac{a_0}{1-r}\gg \gg \frac{1}{1-\frac{1}{5}}\gg \gg \frac{5}{4}$$

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(b)

Split numerator and obtain two geometric series:

$$\sum _{n=0}^{\infty }\frac{2+3^n}{5^n}\gg \gg \sum _{n=0}^{\infty }2{\left(\frac{1}{5}\right)}^n+\sum _{n=0}^{\infty }{\left(\frac{3}{5}\right)}^n$$

Geometric series total sum formula:

$$\gg \gg \frac{2}{1-\frac{1}{5}}+\frac{1}{1-\frac{3}{5}}\gg \gg 5$$Link to original

03

Geometric series

Compute the following summation values using the sum formula for geometric series.

(a) ${\displaystyle \sum _{n=-4}^{\infty }{(-\frac{4}{9})}^n}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }{e}^{3-2n}}$

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Solution

Solutions - 0150-03

(a)

First term:

$$a_{-4}={(-\frac{4}{9})}^{-4}\gg \gg \frac{3^8}{2^8}$$

Common ratio is $r=-4/9$.

Geometric series summation formula, always first term on top:

$$\begin{array}{c}S=\frac{a_{-4}}{1-r}\gg \gg \frac{3^8}{2^8}\left(\frac{1}{1--4/9}\right)\\ \\ \gg \gg \frac{3^8}{2^8}\frac{9}{13}\gg \gg \frac{3^{10}}{2^8(13)}\gg \gg \approx 17.74\end{array}$$

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(b) ${\sum }_{n=0}^{\infty }{e}^{3-2n}$

First term: $a_0=e^3$

Common ratio:

$$r=\frac{e^{3-2(n+1)}}{e^{3-2n}}\gg \gg \frac{e\cdot e^{-2n}}{e^3\cdot e^{-2n}}\gg \gg \frac{1}{e^2}$$

Geometric series summation formula:

$$S=\frac{a_0}{1-r}\gg \gg \frac{e^3}{1-\frac{1}{e^2}}\gg \gg \frac{e^5}{e^2-1}$$Link to original

Positive series

05

IT, DCT, LCT

Determine whether the series converges by checking applicability and then applying the designated convergence test.

(a) Integral Test: ${\displaystyle \sum _{n=2}^{\infty }\frac{\ln n}{n^2}}$

(b) Direct Comparison Test: ${\displaystyle \sum _{n=1}^{\infty }\frac{n^3}{n^5+4n+1}}$

(c) Limit Comparison Test: ${\displaystyle \sum _{n=2}^{\infty }\frac{n^2}{n^4-1}}$

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Solution

Solutions - 0180-05

(a)

Set $f(x)={\displaystyle \frac{\ln x}{x^2}}$. Applicability of the IT:

• $f(x)$ is is continuous other than at $x=0$, and the series starts at $1$.
• $f(x)>0$ since $\ln x>0$ for all $x>1$, and $x^2>0$ for all $x\ge 1$.
• $f^{\prime }(x)={\displaystyle \frac{x-2x\ln x}{x^3}}$. This is zero at $x=\sqrt{e}$. For $x>\sqrt{e}$, $f^{\prime }(x)<0$, so the function is decreasing.

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Apply the integral test:

$$\begin{array}{c}{\int }_2^{\infty }\frac{\ln x}{x^2}dx=\underset{R\to \infty }{\lim }{\int }_2^R\frac{\ln x}{x^2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{[-\frac{\ln x}{x}-\frac{1}{x}]|}_2^R\\ \\ \gg \gg (0-0)-(-\frac{\ln 2}{2}--\frac{1}{2})\gg \gg \frac{1+\ln 2}{2}\end{array}$$

This is finite, so the original series converges by the IT.

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(b) For very large $n$, the large powers dwarf the small powers, and the terms look like ${\displaystyle \frac{n^3}{n^5}}$ which equals ${\displaystyle \frac{1}{n^2}}$. So we take this for a comparison series and apply the DCT:

$$\frac{n^3}{n^5+4n+1}<\frac{n^3}{n^5}=\frac{1}{n^2}$$

But ${\displaystyle \sum \frac{1}{n^2}}$ converges ($p=2>1$). So by the DCT, the original series converges.

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(c) For very large $n$, the large powers dwarf the small powers, and the terms look like ${\displaystyle \frac{n^2}{n^4}}$ which equals ${\displaystyle \frac{1}{n^2}}$. So we take this for a comparison series and apply the LCT:

$$a_n\cdot b_n^{-1}=\frac{n^2}{n^4-1}\cdot \frac{n^2}{1}\gg \gg \frac{n^4}{n^4-1}\stackrel{n\to \infty }{⟶}1=L$$

Observe that $0<L<\infty$. Therefore the LCT says that both series converge or both diverge.

We know that ${\displaystyle \sum \frac{1}{n^2}}$ converges ($p=2>1$). So by the LCT, the original series converges.

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03

Limit Comparison Test (LCT)

Use the Limit Comparison Test to determine whether the series converges:

$${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+\ln n}}$$

Show your work. You must check that the test is applicable.

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Solution

Solutions - 0180-03

The series has positive terms.

Notice that $\sqrt{n}$ is much greater than $\ln n$ for large $n$. So we anticipate that this term will dominate, and we compare the series to ${\displaystyle \frac{1}{\sqrt{n}}}$.

$$\begin{array}{c}{a}_n\cdot b_n^{-1}\gg \gg \frac{1}{\sqrt{n}+\ln n}\cdot \sqrt{n}\\ \\ \gg \gg \frac{n^{-1/2}}{n^{-1/2}}\cdot \frac{n^{1/2}}{n^{1/2}+\ln n}\gg \gg \frac{1}{1+{\displaystyle \frac{\ln n}{n^{1/2}}}}\end{array}$$

We seek the limit of this as $n\to \infty$. Apply L’Hopital’s Rule to the fraction:

$$\frac{\ln x}{x^{1/2}}{\underset{d/dx}{⟶}}^{d/dx}\frac{1/x}{\frac{1}{2\sqrt{x}}}\gg \gg \frac{2}{\sqrt{x}}\stackrel{x\to \infty }{⟶}0$$

Therefore, by continuity:

$$\frac{1}{1+{\displaystyle \frac{\ln n}{n^{1/2}}}}\stackrel{n\to \infty }{⟶}\frac{1}{1+0}\gg \gg 1=L$$

Since $0<L<\infty$, the LCT says that both series converge or diverge.

Since ${\displaystyle \sum \frac{1}{\sqrt{n}}}$ diverges ($p=1/2$), the original series must diverge.

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Alternating series

01

Absolute and conditional convergence

Determine whether the series are absolutely convergent, conditionally convergent, or divergent.

Show your work. You must check applicability of tests.

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^{1/3}}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^n{n}^4}{n^3+1}}$

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Solution

Solutions - 0190-01

(a)

$$\begin{array}{c}\left|\frac{{(-1)}^{n-1}}{n^{1/3}}\right|\gg \gg \frac{1}{n^{1/3}}\\ \\ \sum \frac{1}{n^{1/3}}\text{diverges: }p=1/3<1\end{array}$$

Therefore it is not absolutely converging. Proceed to the AST.

• Passes SDT? Yes, $1/n^{1/3}\to 0$.
• Decreasing? Yes, denominator is increasing.

Therefore the AST applies and says this converges. So it converges conditionally.

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(b)

$$\left|\frac{{(-1)}^n{n}^4}{n^3+1}\right|\gg \gg \frac{n^4}{n^3+1}\to \infty$$

So this fails the SDT, hence it diverges.

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03

Alternating series: error estimation

Find the approximate value of ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n!}}$ such that the error $E_n$ satisfies $|E_n|<0.005$.

How many terms are needed?

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Solution

Solutions - 0190-03

We use the alternating series test error bound formula. (AKA: “Next Term Bound”)

$$|E_n|\le a_{n+1}$$

We seek the smallest $n$ such that $a_{n+1}<0.005$. What that happens, we will have:

$$|E_n|\le a_{n+1}<0.005\gg \gg |E_n|<0.005$$

Our formula for $a_n$:

$$a_n=\frac{1}{n!}\gg \gg a_{n+1}=\frac{1}{(n+1)!}$$

We cannot easily solve for $n$ to provide $a_{n+1}<0.005$, so we just start listing out the terms:

$$\begin{array}{c}{a}_1=1,a_2=0.5,a_3=0.167,\\ \\ a_4=0.0417,a_5=0.00833,a_6=0.00139\end{array}$$

We see that $a_6$ is the first term less than 0.005, so $n+1=6$ and we need the first 5 terms:

$$\sum _{n=1}^5\frac{{(-1)}^{n-1}}{n!}=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!}\gg \gg \approx 0.633$$Link to original

Ratio test and Root test

02

Ratio and root tests

Apply the ratio test or the root test to determine whether each of the following series is absolutely convergent, conditionally convergent, or divergent.

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-2)}^n}{n^{100}}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{5n}{10n+4}\right)}^n}$ $$ (c) ${\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{3^n}}$

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Solution

Solutions - 0200-02

(a)

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|\gg \gg \frac{2^{n+1}}{{(n+1)}^{100}}\cdot \frac{n^{100}}{2^n}\\ \\ \gg \gg 2{\left(\frac{n}{n+1}\right)}^{100}\stackrel{n\to \infty }{⟶}2=L>1\end{array}$$

Since $L>1$, the ratio test says that the series diverges.

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(b)

$$\sqrt[n]{|a_n|}=\frac{5n}{10n+4}\stackrel{n\to \infty }{⟶}\frac{5}{10}=L<1$$

Since $L<1$, the root test says that the series converges.

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(c)

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=\frac{\sqrt{n+1}}{3^{n+1}}\cdot \frac{3^n}{\sqrt{n}}\\ \\ \gg \gg \frac{1}{3}\sqrt{\frac{n+1}{n}}\stackrel{n\to \infty }{⟶}\frac{1}{3}=L<1\end{array}$$

Since $L<1$, the ratio test says that the series converges.

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Power series: Radius and Interval

02

Power series - radius and interval

Find the radius and interval of convergence for these power series:

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(x+3)}^n}{n!}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^n\frac{{(x-7)}^n}{n}}$ $$ (c) ${\displaystyle \sum _{n=12}^{\infty }{n}^n{(x-2)}^n}$

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Solution

Solutions - 0220-02

(a)

$$\begin{array}{c}|\frac{{(x+3)}^{n+1}}{(n+1)!}\cdot \frac{n!}{{(x+3)}^n}|\gg \gg \frac{1}{n+1}|x+3|\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore $R=\infty$ and $I=(-\infty ,+\infty )$.

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(b)

$$|\frac{{(x-7)}^{n+1}}{n+1}\cdot \frac{n}{{(x-7)}^n}|\gg \gg \frac{n}{n+1}|x-7|\stackrel{n\to \infty }{⟶}0$$

Therefore $R=1$ and the preliminary interval is $(\mathrm{6,8})$.

At $x=6$ we have ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(-1)}^n}{n}=\sum _{n=0}^{\infty }\frac{1}{n}}$. This diverges ($p=1$).

At $x=8$ we have ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(+1)}^n}{n}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n}}$. This converges by the AST.

Therefore, the final interval of convergence is $I=(\mathrm{6,8}]$.

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(c)

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

Observe that $1+\frac{1}{n}>1$ so ${\lim }_{n\to \infty }{(1+\frac{1}{n})}^n\ge 1$. Assume $x\ne 2$. Then:

$$(n+1){(1+\frac{1}{n})}^n|x-2|\stackrel{n\to \infty }{⟶}\infty$$

If $x=2$, then of course the series is $\sum 0$ and converges to $0$.

Therefore $R=0$ and $I=\{2\}$.

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Series tests: strategy tips

02

Various limits, Part II

Find the limits. You may use $+\infty$ or $-\infty$ or $\mathrm{DNE}$ as appropriate. Braces indicate sequences.

• C = Convergent
• AC = Absolutely Convergent
• CC = Conditionally Convergent
• D = Divergent

$a_n$	${\lim }_{n\to \infty }{a}_n$	$\left\{a_n\right\}$ C or D	${\lim }_{n\to \infty }{(-1)}^n{a}_n$	$\left\{{(-1)}^n{a}_n\right\}$ C or D	$\sum a_n$ AC, CC, or D	$\sum {(-1)}^n{a}_n$ AC, CC, or D
${\displaystyle \frac{4n!}{2^n}}$
${\displaystyle \frac{(n+2)3^n}{n!}}$
${\displaystyle \frac{4^n}{{(3n)}^n}}$
${\displaystyle \frac{1}{(2n+1)!}}$

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Solution

Solutions - 0210-02

$a_n$	${\lim }_{n\to \infty }{a}_n$	$\left\{a_n\right\}$ C or D	${\lim }_{n\to \infty }{(-1)}^n{a}_n$	$\left\{{(-1)}^n{a}_n\right\}$ C or D	$\sum a_n$ AC, CC, or D	$\sum {(-1)}^n{a}_n$ AC, CC, or D
${\displaystyle \frac{4n!}{2^n}}$	$\infty$	$D$	$DNE$	$D$	$D$	$D$
${\displaystyle \frac{(n+2)3^n}{n!}}$	0	$C$	0	$C$	$AC$	$AC$
${\displaystyle \frac{4^n}{{(3n)}^n}}$	0	$C$	0	$C$	$AC$	$AC$
${\displaystyle \frac{1}{(2n+1)!}}$	0	$C$	0	$C$	$AC$	$AC$

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Power series as functions

01

Modifying geometric power series

Consider the geometric power series $\frac{1}{1-x}=1+x+x^2+x^3+\cdots ={\sum }_{n=0}^{\infty }{x}^n$ for $|x|<1$.

For this problem, you should modify the series for $\frac{1}{1-x}$.

(a) Write ${\displaystyle \frac{1}{5-x}}$ as a power series and determine its interval of convergence.

(b) Write ${\displaystyle \frac{1}{16+2x^3}}$ as a power series and determine its interval of convergence.

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Solution

Solutions - 0230-01

(a)

$$\begin{array}{c}\frac{1}{5-x}\gg \gg \frac{1}{5}\cdot \frac{1}{(1-\frac{x}{5})}\\ \\ \gg \gg \frac{1}{5}\sum _{n=0}^{\infty }{\left(\frac{x}{5}\right)}^n\gg \gg \sum _{n=0}^{\infty }\frac{1}{5^{n+1}}{x}^n\end{array}$$

The geometric series for $\frac{1}{1-x}$ converges when $|x|<1$ and diverges for $|x|\ge 1$. So ours will converge when $\left|{\displaystyle \frac{x}{5}}\right|<1$, which is when $|x|<5$, and diverge otherwise. The interval is therefore $(-5,+5)$.

One can check this in more detail by doing the ratio test:

$$|\frac{x^{n+1}}{5^{n+1}}\cdot \frac{5^n}{x^n}|\gg \gg \frac{1}{5}|x|$$

But we must be careful: the ratio test will not tell us what happens at the endpoints of the interval. If we apply the ratio test here, we would have to check the endpoint separately. But if we use the known result for geometric series, we know it diverges at both endpoints.

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(b)

$$\begin{array}{c}\frac{1}{16+2x^3}\gg \gg \frac{1}{16}\cdot \frac{1}{1-(-{\displaystyle \frac{x^3}{8}})}\\ \\ \gg \gg \frac{1}{16}\sum _{n=0}^{\infty }{(-\frac{x^3}{8})}^n\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{2^{3n+4}}{x}^{3n}\end{array}$$

The geometric series for $\frac{1}{1-x}$ converges when $|x|<1$. So our series will converge when $\left|{\displaystyle \frac{x^3}{8}}\right|<1$, which is when $|x|<2$, and diverges for $|x|\ge 2$. So the interval is $(-\mathrm{2,2})$.

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03

Finding a power series

Find a power series representation for these functions:

(a) $f(x)={\displaystyle \frac{x^2}{x^4+81}}$ $$ (b) $g(x)=x^2\ln (1+x)$

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Solution

Solutions - 0230-03

(a)

$$\begin{array}{c}\frac{x^2}{x^4+81}\gg \gg \frac{x^2}{81}\cdot \frac{1}{1-(-\frac{x^4}{81})}\\ \\ \gg \gg \frac{x^2}{81}\sum _{n=0}^{\infty }{(-\frac{x^4}{81})}^n\gg \gg f(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{3^{4n+4}}{x}^{4n+2}\end{array}$$

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Another approach:

$$\begin{array}{ccc}& \frac{x^2}{x^4+81}\gg \gg \frac{x^2}{a}\frac{a}{{\left(x^2\right)}^2+a^2},a=9& \text{(A)}\end{array}$$

We know that:

$$\begin{array}{c}\frac{a}{u^2+a^2}=\frac{d}{du}{\tan }^{-1}\left(\frac{u}{a}\right)\gg \gg \frac{d}{du}\sum _{n=0}^{\infty }{(-1)}^n\frac{{(u/a)}^{2n+1}}{2n+1}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{u}{a}\right)}^{2n}\frac{1}{a}\end{array}$$

Plug in $u=x^2$:

$$\gg \gg \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{x^2}{a}\right)}^{2n}\frac{1}{a}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{a^{2n+1}}{x}^{4n}$$

Complete:

$$\begin{array}{c}\text{A:}\frac{x^2}{a}\frac{a}{{\left(x^2\right)}^2+a^2}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{a^{2n+2}}{x}^{4n+2}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{9^{2n+2}}{x}^{4n+2}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{3^{4n+4}}{x}^{4n+2}\end{array}$$

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(b)

Notice:

$$\frac{d}{dx}\ln (1+x)=\frac{1}{1+x}\gg \gg \sum _{n=0}^{\infty }{(-x)}^n$$

Integrate:

$$\ln (1+x)=\int \sum _{n=0}^{\infty }{(-x)}^{n}dx\gg \gg C+\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+1}$$

Plug in $x=0$ to solve and find $C=0$.

Now then:

$$\begin{array}{c}{x}^2\ln (1+x)\gg \gg x^2\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+1}\\ \\ \gg \gg g(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+3}\end{array}$$Link to original

Taylor and Maclaurin series

01

Maclaurin series

For each of these functions, find the Maclaurin series, and the interval on which the expansion is valid.

(a) $x\ln (1-5x)$ $$ (b) $x^2\cos (x^3)$

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Solution

Solutions - 0240-01

(a)

$$\begin{array}{c}\ln (1-u)=\sum _{n=0}^{\infty }-\frac{u^{n+1}}{n+1}\\ \\ \gg \gg \ln (1-5x)=\sum _{n=0}^{\infty }-\frac{{(5x)}^{n+1}}{n+1}\\ \\ \gg \gg x\ln (1-5x)=\sum _{n=0}^{\infty }-\frac{5^{n+1}}{n+1}{x}^{n+2}\end{array}$$

$I=[-1/5,+1/5)$

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(b)

$$\begin{array}{c}\cos (u)=\sum _{n=0}^{\infty }{(-1)}^n\frac{u^{2n}}{(2n)!}\\ \\ \gg \gg \cos (x^3)=\sum _{n=0}^{\infty }{(-1)}^n\frac{{(x^3)}^{2n}}{(2n)!}\\ \\ \gg \gg x^2\cos (x^3)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{(2n)!}{x}^{6n+2}\end{array}$$

$I=(-\infty ,\infty )$

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05

Discovering the function from its Maclaurin series

For each of these series, identify the function of which it is the Maclaurin series, and evaluate the function at an appropriate choice of $x$ to find the total sum for the series.

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+1}}{4^{2n+1}(2n+1)!}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{2^{2n}}{n!}}$ $$ (c) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+2}}{3^{2n+1}(2n)!}}$

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Solution

Solutions - 0240-05

(a)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+1}}{4^{2n+1}(2n+1)!}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{{(\pi /4)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sin x{|}_{x=\pi /4}\gg \gg \frac{\sqrt{2}}{2}\end{array}$$

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(b)

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{2^{2n}}{n!}\gg \gg \sum _{n=0}^{\infty }\frac{{(4)}^n}{n!}\\ \\ \gg \gg e^x{|}_{x=4}\gg \gg e^4\end{array}$$

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(c)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+2}}{3^{2n+1}(2n)!}\gg \gg \frac{{\pi }^2}{3}\sum _{n=0}^{\infty }{(-1)}^n\frac{{(\pi /3)}^{2n}}{(2n)!}\\ \\ \gg \gg \frac{{\pi }^2}{3}\cos x{|}_{x=\pi /3}\gg \gg \frac{{\pi }^2}{6}\end{array}$$Link to original

06

Summing a Maclaurin series by guessing its function

For each of these series, identify the function of which it is the Maclaurin series:

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{5x^{4n+2}}{(2n+1)!}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-5x)}^{n+1}}{n+1}}$

Now find the total sums for these series:

(c) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-5)}^n}{n!}}$ $$ (d) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{9^n(2n)!}}$

(Hint: for (c)-(d), do the process in (a)-(b), then evaluate the resulting function somewhere.)

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Solution

Solutions - 0240-06

(a)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{5x^{4n+2}}{(2n+1)!}\gg \gg 5\sum _{n=0}^{\infty }{(-1)}^n\frac{{(x^2)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg 5\sin (x^2)\end{array}$$

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(b)

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-5x)}^{n+1}}{n+1}\gg \gg -\sum _{n=0}^{\infty }-\frac{{(-5x)}^{n+1}}{n+1}\\ \\ \gg \gg -\ln (1-(-5x))\gg \gg -\ln (1+5x)\end{array}$$

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(c)

$$\sum _{n=0}^{\infty }\frac{{(-5)}^n}{n!}\gg \gg e^x{|}_{x=-5}\gg \gg e^{-5}$$

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(d)

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{9^n(2n)!}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{{(\pi /3)}^{2n}}{(2n)!}\\ \\ \gg \gg \cos x{|}_{x=\pi /3}\gg \gg \cos \frac{\pi }{3}\gg \gg \frac{1}{2}\end{array}$$Link to original

09

Large derivative at $x=0$ using pattern of Maclaurin series

Consider the function $f(x)=x^2\sin \left(5x^3\right)$. Find the value of $f^{(35)}(0)$.

(Hint: find the rule for coefficients of the Maclaurin series of $f(x)$ and then plug in $0$.)

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Solution

Solutions - 0240-09

$$\begin{array}{c}{x}^2\sin \left(5x^3\right)=x^2\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(5x^3\right)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{5^{2n+1}}{(2n+1)!}{x}^{6n+5}\end{array}$$

The Coefficient-Derivative Identity says that $f^{(k)}(0)={\displaystyle \frac{a_k}{k!}}$ where $a_k$ is the coefficient of the $k^{\text{th}}$ power $x^k$.

Solve:

$$x^{35}=x^{6n+5}\gg \gg 35=6n+5\gg \gg n=5$$

So, for the coefficient $a_{35}$:

$$\begin{array}{c}{(-1)}^5\frac{5^{2\cdot 5+1}}{(2\cdot 5+1)!}\gg \gg -\frac{5^{11}}{11!}\\ \\ \gg \gg f^{(35)}(0)=35!a_{35}\gg \gg -5^{11}\frac{35!}{11!}\end{array}$$Link to original

Applications of Taylor series

01

Approximating $1/e$

Using the series representation of $e^x$, show that:

$$\frac{1}{e}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots$$

Now use the alternating series error bound to approximate $\frac{1}{e}$ to an error within ${10}^{-3}$.

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Solution

Solutions - 0250-01

Notice that ${\displaystyle \frac{1}{e}}=e^{-1}$.

We have this series for $e^x$:

$$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$

Therefore:

$$\begin{array}{c}{e}^{-1}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}\\ \\ \gg \gg \frac{1}{e}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots \end{array}$$

Now we use the “Next Term Bound” rule. Calculate terms until we find a term less than ${10}^{-3}$:

$$\begin{array}{c}\frac{1}{4!}=4.2\times {10}^{-2},\frac{1}{5!}=8.3\times {10}^{-3},\\ \\ \frac{1}{6!}=1.4\times {10}^{-3},\frac{1}{7!}=2.4\times {10}^{-4}\end{array}$$

So we take the following partial sum approximation:

$$\frac{1}{e}\approx \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}=0.3681$$Link to original

03

Some estimates using series

Find an infinite series representation of ${\displaystyle {\int }_0^1\sin (x^2)dx}$ and then use your series to estimate this integral to within an error of ${10}^{-3}$.

(Use the error bound formula for alternating series.)

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Solution

Solutions - 0250-03

Write the series of the integrand:

$$\begin{array}{c}\sin \left(x^2\right)=\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(x^2\right)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+2}}{(2n+1)!}\end{array}$$

Integrate:

$$\begin{array}{c}{\int }_0^1\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+2}}{(2n+1)!}dx\\ \\ \gg \gg {\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+3}}{(2n+1)!(4n+3)}|}_0^1\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{(2n+1)!(4n+3)}\\ \\ \gg \gg \frac{1}{3}-\frac{1}{3!(7)}+\frac{1}{5!(11)}-\cdots \end{array}$$

Now apply the “Next Term Bound” and look for the first term below ${10}^{-3}$:

$$\frac{1}{3!(7)}=2.4\times {10}^{-2},\frac{1}{5!(11)}=7.6\times {10}^{-4}$$

So we simply add the first two terms:

$$\frac{1}{3}-\frac{1}{3!(7)}\approx 0.3095$$Link to original