W09 Stepwise

Due date: Thursday 3/12, 11:59pm

Positive series

01

01

Integral Test (IT)

Use the Integral Test to determine whether the series converges:

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

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

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Solution

Solutions - 0180-01

$$a_n=\frac{1}{n^2+1}\gg \gg a_x=\frac{1}{x^2+1}$$

First note that:

1. $a_x$ is continuous
2. $a_x$ is positive
3. $a_x$ is monotone decreasing because $x^2+1$ is increasing

Then:

$$\begin{array}{c}{\int }_1^{\infty }\frac{1}{x^2+1}dx\gg \gg \underset{R\to \infty }{\lim }{\int }_1^R\frac{1}{x^2+1}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{\tan }^{-1}(x){|}_1^R\\ \\ \gg \gg \underset{R\to \infty }{\lim }{\tan }^{-1}(R)-\frac{\pi }{4}\\ \\ \gg \gg \frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\end{array}$$

Since this is finite, the integral test establishes that the series converges.

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02

02

Direct Comparison Test (DCT)

Determine whether the series is convergent by using the Direct Comparison Test.

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

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^{1/3}+2^n}}$ $$ (b) ${\displaystyle \sum _{k=2}^{\infty }\frac{\sqrt{k}}{k-1}}$

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Solution

Solutions - 0180-02

(a) The series has positive terms.

$$\frac{1}{n^{1/3}+2^n}<\frac{1}{2^n}$$

But ${\displaystyle \sum \frac{1}{2^n}}$ converges because it is geometric with $r=1/2$. By the DCT, the original series must converge.

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(b) The series has positive terms.

$$\frac{\sqrt{k}}{k-1}>\frac{\sqrt{k}}{k}=\frac{1}{k^{1/2}}$$

But ${\displaystyle \sum \frac{1}{k^{1/2}}}$ diverges because it is a $p$-series with $p=1/2<1$. By the DCT, the original series must diverge.

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03

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

04

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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05

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