W09 - Notes

Simple divergence test

Videos

Review Videos

Videos, Math Dr. Bob

• Geometric series and SDT, again: Geometric series, Simple Divergence Test (aka “Limit Test”)
• Integral test: Basics
• Integral test: $p$-series
  • Extra: Integral test: Further examples
  • Extra: Integral test: Estimations

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01 Theory

Theory 1

Simple Divergence Test (SDT)

Applicability: Any series.

Test Statement:

$$\underset{n\to \infty }{\lim }{a}_n\ne 0⟹\sum _{n=1}^{\infty }{a}_n\text{diverges}$$

AKA the “Not Even Close” test

The converse is not valid. For example, ${\sum }_{n=1}^{\infty }\frac{1}{n}$ diverges even though ${\lim }_{n\to \infty }\frac{1}{n}=0$.

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02 Illustration

Example - Simple Divergence Test: examples

Simple divergence test: examples

(a) Consider: ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{4n+1}}$

This diverges by the SDT because $a_n\to \frac{1}{4}$ and not $0$.

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(b) Consider: ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{n}{n+1}}$

This diverges by the SDT because ${\lim }_{n\to \infty }{a}_n=\text{DNE}$.

We can say the terms “converge to the pattern $+1,-1,+1,-1,\dots$,” but that is not a limit value.

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Positive series

Videos

Review Videos

• Direct Comparison Test: Theory and basic examples
• Direct Comparison Test: Series $\frac{1}{\ln n}$
• Limit Comparison Test: Theory and basic examples
• Limit Comparison Test: Further examples

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03 Theory

Theory 1

Positive series

A series is called positive when its individual terms are positive, i.e. $a_n>0$ for all $n$.

The partial sum sequence $S_N$ is monotone increasing for a positive series.

By the monotonicity test for convergence of sequences, $S_N$ therefore converges whenever it is bounded above. If $S_N$ is not bounded above, then ${\sum }_{n=1}^{\infty }{a}_n$ diverges to $+\infty$.

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Another test, called the integral test, studies the terms of a series as if they represent rectangles with upper corner pinned to the graph of a continuous function.

To apply the test, we must convert the integer index variable $n$ in $a_n$ into a continuous variable $x$. This is easy when we have a formula for $a_n$ (provided it doesn’t contain factorials or other elements dependent on integrality).

Integral Test (IT)

Applicability: $f(x)$ must be:

• Continuous
• Positive
• Monotone decreasing

Test Statement:

$$\sum _{n=1}^{\infty }{a}_n\text{converges}⟺{\int }_1^{\infty }f(x)dx\text{converges}$$

Extra - Integral test: explanation

To show that integral convergence implies series convergence, consider the diagram:

This shows that ${\sum }_{n=2}^N{a}_n\le {\int }_1^{N}f(x)dx$ for any $N$. Therefore, if ${\int }_1^{\infty }f(x)dx$ converges, then ${\int }_1^{N}f(x)dx$ is bounded (independent of $N$) and so ${\sum }_{n=2}^N{a}_n$ is bounded by that inequality. But ${\sum }_{n=2}^N{a}_n=S_N-a_1$; so by adding $a_1$ to the bound, we see that $S_N$ itself is bounded, which implies that ${\sum }_{n=1}^{\infty }{a}_n$ converges.

To show that integral divergence implies series divergence, consider a similar diagram:

This shows that ${\sum }_{n=1}^{N-1}{a}_n\ge {\int }_1^{N}f(x)dx$ for any $N$. Therefore, if ${\int }_1^{\infty }f(x)dx$ diverges, then ${\int }_1^{N}f(x)dx$ goes to $+\infty$ as $N\to \infty$, and so ${\sum }_{n=1}^{N-1}{a}_n$ goes to $+\infty$ as well. So ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

Notice: the picture shows $f(x)$ entirely above (or below) the rectangles. This depends upon $f(x)$ being monotone decreasing, as well as $f(x)>0$. (This explains the applicability conditions.)

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Next we use the integral test to evaluate the family of $p$-series, and later we can use $p$-series in comparison tests without repeating the work of the integral test.

$p$-series

A $p$-series is a series of this form: ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$

Convergence properties:

$$p>1:\text{series converges}p\le 1:\text{series diverges}$$

Extra - Proof of $p$-series convergence

(1) To verify the convergence properties of $p$-series, apply the integral test:

Applicability: verify it’s continuous, positive, decreasing.

Convert $n$ to $x$ to obtain the function $f(x)=\frac{1}{x^p}$.

Indeed $\frac{1}{x^p}$ is continuous and positive and decreasing as $x$ increases.

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(2) Apply the integral test.

Integrate, assuming $p\ne 1$:

$$\begin{array}{cc}{\int }_1^{\infty }\frac{1}{x^p}dx& \gg \gg \underset{R\to \infty }{\lim }{\frac{x^{p-1}}{p-1}|}_1^R\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }(\frac{R^{-p+1}}{-p+1}-\frac{1^{-p+1}}{-p+1})\end{array}$$

When $p>1$ we have ${\lim }_{R\to \infty }\frac{R^{-p+1}}{-p+1}=0$

When $p<1$ we have ${\lim }_{R\to \infty }\frac{R^{-p+1}}{-p+1}=\infty$

When $p=1$, integrate a second time:

$$\begin{array}{cc}{\int }_1^{\infty }\frac{1}{x}dx& \gg \gg \underset{R\to \infty }{\lim }\ln x{|}_1^R\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\ln R-\ln 1\gg \gg \infty \end{array}$$

Conclude: the integral converges when $p>1$ and diverges when $p\le 1$.

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We could instead immediately refer to the convergence results for $p$-integrals instead of reproving them here.

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04 Illustration

Example - $p$-series examples

p-series examples

By finding $p$ and applying the $p$-series convergence properties:

We see that ${\sum }_{n=1}^{\infty }\frac{1}{n^{1.1}}$ converges: $p=1.1$ so $p>1$

But ${\sum }_{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges: $p=1/2$ so $p\le 1$

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Example - Integral test - pushing the envelope of convergence

Integral test - pushing the envelope of convergence

Does ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\ln n}}$ converge?

Does ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n{(\ln n)}^2}}$ converge?

Notice that $\ln n$ grows very slowly with $n$, so $\frac{1}{n\ln n}$ is just a little smaller than $\frac{1}{n}$ for large $n$, and similarly $\frac{1}{n{(\ln n)}^2}$ is just a little smaller still.

Solution

(1) The two series lead to the two functions $f(x)=\frac{1}{x\ln x}$ and $g(x)=\frac{1}{x{(\ln x)}^2}$.

Check applicability.

Clearly $f(x)$ and $g(x)$ are both continuous, positive, decreasing functions on $x\in [2,\infty )$.

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(2) Apply the integral test to $f(x)$.

Integrate $f(x)$:

$$\begin{array}{cc}{\int }_2^{\infty }\frac{1}{x\ln x}dx& \gg \gg {\int }_{u=\ln 2}^{\infty }\frac{1}{u}du\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\ln u{|}_{\ln 1}^R\gg \gg \infty \end{array}$$

Conclude: ${\sum }_{n=2}^{\infty }\frac{1}{n\ln n}$ diverges.

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(3) Apply the integral test to $g(x)$.

Integrate $g(x)$:

$$\begin{array}{cc}{\int }_2^{\infty }\frac{1}{x{(\ln x)}^2}dx& \gg \gg {\int }_{u=\ln 2}^{\infty }\frac{1}{u^2}du\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }-u^{-1}{|}_{\ln 2}^R\gg \gg \frac{1}{\ln 2}\end{array}$$

Conclude: ${\sum }_{n=2}^{\infty }\frac{1}{n{(\ln n)}^2}$ converges.

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05 Theory

Theory 2

Direct Comparison Test (DCT)

Applicability: Both series are positive: $a_n>0$ and $b_n>0$.

Test Statement: Suppose $a_n\le b_n$ for large enough $n$. (Meaning: for $n\ge N$ with some given $N$.) Then:

• Smaller pushes up bigger:

$$\sum _{n=1}^{\infty }{a}_n\text{diverges}⟹\sum _{n=1}^{\infty }{b}_n\text{diverges}$$

• Bigger controls smaller:

$$\sum _{n=1}^{\infty }{b}_n\text{converges}⟹\sum _{n=1}^{\infty }{a}_n\text{converges}$$

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06 Illustration

Example - Direct comparison test: rational functions

Direct comparison test: rational functions

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}{3}^n}}$

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) ${\displaystyle \sum _{n=1}^{\infty }\frac{{\cos }^{2}n}{n^3}}$

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) ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{n^3+1}}$

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

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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07 Theory

Theory 3

Some series can be compared using the DCT after applying certain manipulations and tricks.

For example, consider the series ${\sum }_{n=2}^{\infty }\frac{1}{n^2-1}$. We suspect convergence because $a_n\approx \frac{1}{n^2}$ for large $n$. But unfortunately, $a_n>\frac{1}{n^2}$ always, so we cannot apply the DCT.

We could make some ad hoc arguments that do use the DCT, eventually:

Trick Method 1:

• Observe that for $n>1$ we have $\frac{1}{n^2-1}\le \frac{10}{n^2}$. (Check it!)
• But $\sum \frac{10}{n^2}$ converges, indeed its value is $10\cdot \sum \frac{1}{n^2}$, which is $\frac{10{\pi }^2}{6}$.
• So the series $\sum \frac{1}{n^2-1}$ converges.

Trick Method 2:

• Observe that we can change the letter $n$ to $n+1$ by starting the new $n$ at $n=1$.
• Then we have:

$$\sum _{n=2}^{\infty }\frac{1}{n^2-1}=\sum _{n=1}^{\infty }\frac{1}{{(n+1)}^2-1}=\sum _{n=1}^{\infty }\frac{1}{n^2+2n}$$

• This last series has terms smaller than $\frac{1}{n^2}$ so the DCT with $b_n=\frac{1}{n^2}$ (a convergent $p$-series) shows that the original series converges too.

These convoluted arguments suggest that a more general version of Comparison is possible.

Indeed, it is sufficient to compare the relative large-n behavior of the two series. We use the termwise ratios to estimate comparative behavior for increasing $n$.

Limit Comparison Test (LCT)

Applicability: Both series are positive: $a_n>0$ and $b_n>0$.

Test Statement: Suppose that ${\lim }_{n\to \infty }{\displaystyle \frac{a_n}{b_n}=L}$. Then: If $0<L<\infty$, i.e. finite non-zero, then:

$$\sum a_n\text{converges}⟺\sum b_n\text{converges}$$

Extra - LCT edge cases

If $L=0$ or $L=\infty$, we can still draw an inference, but only in one direction:

• If $L=0$:

$$\sum b_n\text{converges}⟹\sum a_n\text{converges}$$

• If $L=\infty$:

$$\sum b_n\text{diverges}⟹\sum a_n\text{diverges}$$

Extra - Limit Comparison Test explanation

Suppose $a_n/b_n\to L$ and $0<L<\infty$. Then for $n$ sufficiently large, we know $a_n/b_n<L+1$.

Doing some algebra, we get $a_n<(L+1)b_n$ for $n$ large.

If $\sum b_n$ converges, then $\sum (L+1)b_n$ also converges (constant multiple), and then the DCT implies that $\sum a_n$ converges.

Conversely: we also know that $b_n/a_n\to 1/L$, so $b_n<(1/L+1)a_n$ for all $n$ sufficiently large. Thus if $\sum a_n$ converges, $\sum (1/L+1)a_n$ also converges, and by the DCT again $\sum b_n$ converges too.

The cases with $L=0$ or $L=\infty$ are handled similarly.

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08 Illustration

Example - Limit Comparison Test examples

Limit comparison test examples

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{2^n-1}}$

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) ${\displaystyle \sum _{n=1}^{\infty }\frac{2n^2+3n}{\sqrt{5+n^5}}}$

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) ${\displaystyle \sum _{n=2}^{\infty }\frac{n^2}{n^4-n-1}}$

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

Videos

Review Videos

Videos, Math Dr. Bob:

• Alternating Series Test: Theory and basic examples
• Alternating Series Test: Remainder estimates
• Alternating Series Test: Further remainder estimates

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09 Theory

Theory 1

Consider these series:

$$\begin{array}{cc}1& +\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots =\infty \\ & \\ -1& -\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\cdots =-\infty \\ & \\ 1& -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\cdots =\ln 2\\ & \\ 1& +\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\cdots =?\end{array}$$

The absolute values of terms are the same between these series, only the signs of terms change.

The first is a positive series because there are no negative terms.

The second series is the negation of a positive series – the study of such series is equivalent to that of positive series, just add a negative sign everywhere. These signs can be factored out of the series. (For example $\sum -\frac{1}{n}=-\sum \frac{1}{n}$.)

The third series is an alternating series because the signs alternate in a strict pattern, every other sign being negative.

The fourth series is not alternating, nor is it positive, nor negative: it has a mysterious or unknown pattern of signs.

A series with any negative signs present, call it ${\sum }_{n=1}^{\infty }{a}_n$, converges absolutely when the positive series of absolute values of terms, namely ${\sum }_{n=1}^{\infty }|a_n|$, converges.

THEOREM: Absolute implies ordinary

If a series ${\sum }_{n=1}^{\infty }{a}_n$ converges absolutely, then it also converges as it stands.

A series might converge due to the presence of negative terms and yet not converge absolutely:

A series ${\sum }_{n=1}^{\infty }{a}_n$ is said to be converge conditionally when the series converges as it stands, but the series produced by inserting absolute values, namely ${\sum }_{n=1}^{\infty }|a_n|$, diverges.

The alternating harmonic series above, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2$, is therefore conditionally convergent. Let us see why it converges. We can group the terms to create new sequences of pairs, each pair being a positive term. This can be done in two ways. The first creates an increasing sequence, the second a decreasing sequence:

$$\begin{array}{cc}\text{increasing from }0\text{:}& (1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{7}-\frac{1}{8})+\cdots \\ & \\ \text{decreasing from }1\text{:}& 1-(\frac{1}{2}-\frac{1}{3})-(\frac{1}{4}-\frac{1}{5})-(\frac{1}{6}-\frac{1}{7})-\cdots \end{array}$$

Suppose $S_N$ gives the sequence of partial sums of the original series. Then $S_{2N}$ gives the first sequence of pairs, namely $S_2$, $S_4$, $S_6$, $\dots$ . And $S_{2N-1}$ gives the second sequence of pairs, namely $S_1$, $S_3$, $S_5$, $\dots$ .

The second sequence shows that $S_N$ is bounded above by $1$, so $S_{2N}$ is monotone increasing and bounded above, so it converges. Similarly $S_{2N-1}$ is monotone decreasing and bounded below, so it converges too, and of course they must converge to the same thing.

The fact that the terms were decreasing in magnitude was an essential ingredient of the argument for convergence. This fact ensured that the parenthetical pairs were positive numbers.

Alternating Series Test (AST) - “Leibniz Test”

Applicability: Alternating series only: ${\sum }_{n=1}^{\infty }{(-1)}^{n-1}{a}_n$ with $a_n>0$

Test Statement: If:

1. $a_n\to 0$ as $n\to \infty$ (i.e. it passes the SDT: if this fails, conclude diverges)
2. $a_n$ are decreasing, so $a_1>a_2>a_3>a_4>\cdots >0$

Then:

$$\sum _{n=1}^{\infty }{(-1)}^{n-1}{a}_n\text{converges}$$

“Next Term Bound” rule for error of the partial sums:

$$|S-S_N|<a_{N+1}$$

Extra - Alternating Series Test: Theory

Just as for the alternating harmonic series, we can form positive paired-up series because the terms are decreasing:

$$\begin{array}{c}(a_1-a_2)+(a_3-a_4)+(a_5-a_6)+\cdots \\ \\ a_1-(a_2-a_3)-(a_4-a_5)-(a_6-a_7)-\cdots \end{array}$$

The first sequence $S_{2N}$ is monotone increasing from $0$, and the second $S_{2N-1}$ is decreasing from $a_1$. The first is therefore also bounded above by $a_1$. So it converges. Similarly, the second converges. Their difference at any point is $S_{2N}-S_{2N-1}$ which is equal to $-a_{2N}$, and this goes to zero. So the two sequences must converge to the same thing.

By considering these paired-up sequences and the effect of adding each new term one after the other, we obtain the following order relations:

$$0<S_2<S_4<S_6<\cdots <S<\cdots <S_5<S_3<S_1=a_1$$

Thus, for any even $2N$ and any odd $2M-1$:

$$S_{2N}<S<S_{2M-1}$$

Now set $M=N$ and subtract $S_{2N-1}$ from both sides:

$$\begin{array}{cc}{S}_{2N}-S_{2N-1}& <S-S_{2N-1}<0\\ & \\ \gg \gg -a_{2N}& <S-S_{2N-1}<0\end{array}$$

Now set $M=N+1$ and subtract $S_{2N}$ from both sides:

$$\begin{array}{cc}0<S-S_{2N}& <S_{2N+1}-S_{2N}\\ & \\ \gg \gg 0<S-S_{2N}& <a_{2N+1}\end{array}$$

This covers both even cases ($n=2N$) and odd cases ($n=2N-1$). In either case, we have:

$$|S-S_n|<a_{n+1}$$

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10 Illustration

Example - Alternating Series Test: Basic illustration

Alternating series test: basic illustration

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{\sqrt{n}}}$ 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) ${\displaystyle \sum _{n=1}^{\infty }\frac{\cos n\pi }{n^2}}$ 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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Example - 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:

$$\begin{array}{cc}& a_{n+1}=\frac{4}{2n+1}<0.001\\ & \\ \gg \gg & \frac{4}{0.001}<2n+1\\ & \\ \gg \gg & 3999<2n\\ & \\ \gg \gg & 2000\le n\end{array}$$

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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