Examples

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 $p=2>1$.

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 $0.001$.