W09 Regular

Due date: Sunday 3/15, 11:59pm

Positive series

01

04

Integral Test (IT)

Determine whether the series is convergent by using the Integral Test.

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

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^{1.1}}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }n{e}^{-n^2}}$ $$ (c) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{n^2}}}$

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Solution

Solutions - 0180-04

(a) Verify applicability of the integral test:

• $f(x)$ is continuous for all $x\ge 1$. (Only discontinuity is at $x=0$, but the series starts at $x=1$.)
• $f(x)\ge 0$ since $x^{1.1}\ge 0$ for all $x\ge 1$.
• $f(x)$ is monotone decreasing, since as $x$ increases, the denominator increases, and the term decreases.

Apply the integral test:

$$\begin{array}{c}{\int }_1^{\infty }\frac{dx}{x^{1.1}}=\underset{R\to \infty }{\lim }{\int }_1^R\frac{dx}{x^{1.1}}\\ \\ \gg \gg \underset{R\to \infty }{\lim }[-10x^{-0.1}]{|}_1^R\gg \gg 0--10=10\end{array}$$

This is finite and the improper integral converges, so the series converges by the Integral Test.

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(b) Verify applicability of the integral test with $f(x)=x{e}^{-x^2}$:

• $f(x)$ is definitely continuous for all $x\in ℝ$.
• $f(x)>0$ since $x>1$ and $e^{-x^2}>0$ for all $x\in ℝ$.
• Decreasing?
  • $f^{\prime }(x)=e^{-x^2}-2x^2{e}^{-x^2}=(1-2x^2)e^{-x^2}$ has zeros at $x=\pm \frac{1}{\sqrt{2}}$.
  • When $x>\frac{1}{\sqrt{2}}$, $f^{\prime }(x)<0$.
  • Series starts at $n=1>\frac{1}{\sqrt{2}}$, so the terms are monotone decreasing.

Apply the integral test:

$$\begin{array}{c}{\int }_1^{\infty }x{e}^{-x^2}dx=\underset{R\to \infty }{\lim }{\int }_1^{R}x{e}^{-x^2}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{[-\frac{e^{-x^2}}{2}]|}_1^R\gg \gg 0--\frac{e^{-1}}{2}\gg \gg \frac{1}{2e}\end{array}$$

This is finite and the improper integral converges, so the series converges by the Integral Test.

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(c) Verify applicability of the integral test for $f(x)=x^{-2/3}$:

• $f(x)$ is continuous for all $x\ge 1$. (The only discontinuity is at $x=0$, but the series starts at $n=1$.)
• $f(x)>0$ since $x^{-2/3}>0$ for all $x>1$.
• $f(x)$ is monotone decreasing, since as $x$ increases, the denominator increases, and the term decreases.

Apply the integral test:

$$\begin{array}{c}{\int }_1^{\infty }\frac{dx}{x^{2/3}}=\underset{R\to \infty }{\lim }{\int }_1^R\frac{dx}{x^{2/3}}\\ \\ \gg \gg \underset{R\to \infty }{\lim }\left({3x^{1/3}|}_1^R\right)\gg \gg \infty -3\end{array}$$

So the improper integral diverges, and the series diverges by the Integral Test.

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02 $★$

06

Limit Comparison Test (LCT)

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

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

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

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Solution

Solutions - 0180-06

For very large $n$, we expect the exponentials to dominate, and the series looks like ${\displaystyle \frac{e^n}{e^{2n}}}={\displaystyle \frac{1}{e^n}}$. This will yield a converging geometric series. Anyway, let us choose $b_n={\displaystyle \frac{1}{e^n}}$ as the comparison series.

$$a_n\cdot b_n^{-1}=\frac{e^n+n}{e^{2n}-n^2}\cdot \frac{e^n}{1}\gg \gg \frac{e^{2n}+n{e}^n}{e^{2n}-n^2}$$

Now divide above and below by the leading power:

$$\begin{array}{c}\frac{e^{2n}+n{e}^n}{e^{2n}-n^2}\gg \gg \frac{e^{-2n}}{e^{-2n}}\cdot \frac{e^{2n}+n{e}^n}{e^{2n}-n^2}\\ \\ \gg \gg \frac{1+n{e}^{-n}}{1-n^2{e}^{-2n}}\end{array}$$

By L’Hopital’s Rule, we find that:

$$\begin{array}{c}n{e}^{-n}\stackrel{n\to \infty }{⟶}0\\ \\ n^2{e}^{-2n}\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore:

$$\frac{1+n{e}^{-n}}{1-n^2{e}^{-2n}}\stackrel{n\to \infty }{⟶}1=L$$

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

Now ${\displaystyle \sum \frac{1}{e^n}=\sum {\left(\frac{1}{e}\right)}^n}$ and this is geometric with $r=\frac{1}{e}<1$. Therefore it converges, and the original series must converge too.

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03

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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Alternating series

04

02

Absolute and conditional convergence

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

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

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

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Solution

Solutions - 0190-07

(a) We first check for absolute convergence:

$$\left|\frac{{(-1)}^n}{1+\frac{1}{n}}\right|\gg \gg \frac{1}{1+\frac{1}{n}}\stackrel{n\to \infty }{⟶}\frac{1}{1+0}=1\ne 0$$

This fails the SDT, so the series diverges!

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(b) Notice that $\cos n\pi ={(-1)}^{n-1}$. Check for absolute convergence:

$$\left|\frac{\cos n\pi }{n^3+1}\right|\gg \gg \frac{1}{n^3+1}$$

Then:

$$\frac{1}{n^3+1}<\frac{1}{n^3}$$

Since ${\displaystyle \sum \frac{1}{n^3}}$ converges ($p=3>1$), the DCT says that ${\displaystyle \sum \frac{1}{n^3+1}}$ converges. So the original series converges absolutely and we are done.

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Sequences and series - additional practice

05

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

06 $★$

07

Limits and convergence

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

(a) ${\displaystyle \ln \left(\frac{2n+1}{3n+4}\right)}$ $$ (b) ${\displaystyle \frac{e^n}{2^n}}$ $$ (c) ${\displaystyle \frac{{(\ln n)}^2}{n}}$ $$ (d) ${\displaystyle \frac{{(-1)}^n{(\ln n)}^2}{n}}$

(e) ${\displaystyle \frac{3-4^n}{2+7\cdot 4^n}}$ $$ (f) ${\displaystyle {(1+\frac{1}{n})}^n}$ $$ (g) ${\displaystyle \frac{1}{\ln (1+\frac{1}{n})}}$

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Solution

Solutions - 0140-07

(a) $\ln (2/3)$

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(b) diverges:

$$\frac{e^n}{2^n}\gg \gg {\left(\frac{e}{2}\right)}^n\stackrel{n\to \infty }{⟶}\infty$$

(Note that $e/2>1$.)

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(c) $0$:

L’Hopital’s Rule:

$$\begin{array}{c}\gg \gg \frac{{(\ln x)}^2}{x}{\underset{d/dx}{⟶}}^{d/dx}\frac{\frac{2}{x}\ln x}{1}=\frac{2\ln x}{x}\\ \\ {\underset{d/dx}{⟶}}^{d/dx}\frac{2/x}{1}=\frac{2}{x}\stackrel{n\to \infty }{⟶}0\end{array}$$

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(d) $0$ by (c), the sign doesn’t affect convergence to $0$

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(e) $-{\displaystyle \frac{1}{7}}$:

Multiply above and below by $4^{-n}$:

$$\begin{array}{c}\frac{3-4^n}{2+7\cdot 4^n}\gg \gg \frac{4^{-n}}{4^{-n}}\cdot \frac{3-4^n}{2+7\cdot 4^n}\\ \\ \gg \gg \frac{-1+\frac{3}{4^n}}{7+\frac{2}{4^n}}\stackrel{n\to \infty }{⟶}-\frac{1}{7}\end{array}$$

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(f) $e$:

This is a well-known formula for $e$. If that formula is not used as the definition of $e$, then it would not be circular reasoning to argue as follows:

$$\begin{array}{c}\underset{n\to \infty }{\lim }{(1+\frac{1}{n})}^n\gg \gg \underset{x\to \infty }{\lim }\exp \left(\ln {(1+\frac{1}{x})}^x\right)\\ \\ \gg \gg \exp (\underset{x\to \infty }{\lim }\ln {(1+\frac{1}{x})}^x)\gg \gg \exp (\underset{x\to \infty }{\lim }\frac{\ln (1+\frac{1}{x})}{1/x})\\ \\ \gg \gg \exp (\underset{x\to \infty }{\lim }\frac{-\frac{1}{x^2}\cdot \frac{1}{1+1/x}}{-1/x^2})\gg \gg \exp (\underset{x\to \infty }{\lim }\frac{-\frac{1}{x^2}\cdot \frac{1}{1+1/x}}{-1/x^2})\\ \\ \gg \gg \exp (1)\end{array}$$

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(g) diverges:

$$\ln (1+\frac{1}{n})\stackrel{n\to \infty }{⟶}\ln (1^{+})=0^{+}$$

So:

$$\frac{1}{\ln (1+\frac{1}{n})}\stackrel{n\to \infty }{⟶}\frac{1}{0^{+}}=+\infty$$Link to original

07 $★$

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

08 $★$

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

09

04

Repeating digits

Using the geometric series formula, find the fractional forms of these decimal numbers:

(a) $0.\bar{2}=0.222222\dots$ $$ (b) $0.4\bar{9}=0.4999999\dots$

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Solution

Solutions - 0150-04

(a)

$$0.2222\dots \gg \gg \frac{2}{10}+\frac{2}{100}+\frac{2}{1000}+\cdots$$

First term: $a_0=2/10$.

Common ratio: $r=1/10$

Geometric series summation formula:

$$S=\frac{a_0}{1-r}\gg \gg \frac{2/10}{1-1/10}\gg \gg \frac{2}{9}$$

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(b)

$$\begin{array}{c}0.49999\dots \gg \gg \frac{4}{10}+\frac{9}{100}+\frac{9}{1000}+\cdots \\ \\ \gg \gg \frac{4}{10}+(\frac{9}{100}+\frac{9}{1000}+\cdots )\end{array}$$

This is geometric starting with $9/100$.

First term: $a_0=9/100$.

Common ratio: $r=1/10$.

Geometric series summation formula:

$$S=\frac{9/100}{1-1/10}\gg \gg \frac{1}{10}$$

Add back the first term:

$$\gg \gg \frac{4}{10}+\frac{1}{10}\gg \gg \frac{1}{2}$$Link to original

10

07

Total area of infinitely many triangles

Find the area of all the triangles as in the figure:

(The first triangle from the right starts at $1$, and going left they never end.)

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Solution

Solutions - 0150-07

Compute the first few areas, with $a_0$ being the area of the largest triangle:

$$\begin{array}{cc}{a}_0& =\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}\\ & \\ a_1& =\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{4}=\frac{1}{16}\\ & \\ a_2& =\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{8}=\frac{1}{32}\end{array}$$

This is a geometric series with $r=\frac{1}{2}$:

$$A=\sum _{n=0}^{\infty }\frac{1}{8}{\left(\frac{1}{2}\right)}^n$$

Geometric series total sum formula:

$$S=\frac{a_0}{1-r}\gg \gg \frac{1/8}{1-1/2}\gg \gg \frac{1}{4}$$Link to original