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• Notes

Alternating Series:

$\sum(-1)^{n-1} a_{n}$

$\lim _{n \rightarrow \infty} a n=0$

$\left\{a_{n}\right\}$is decreasing

let $S=$ sum of $\sum_{n=1}^{\infty}(-1)^{n-1} a n$

$S=a_{1}-a_{2}+a_{3}-a_{4}+\dots(-1)^{n-1} a n+\cdots$

$R n=S-S n$

Error $=|R n|=|S-S n|=|S n-S|$

The error $|R n|$ must be $\leq a_{n+1}$

How many terms of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}$ we need to add in order to find the sum with $| \text { error } |<0.01$

$a_{n}=\frac{1}{n^{2}} \Rightarrow \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=\frac{1}{\infty^{2}}=0$

$f(x)=\frac{1}{n^{2}} \Rightarrow f^{1}(x)=\frac{-2 n}{n^{4}}=\frac{-2}{n^{3}}<0$

$\{a\}$ is decreasing

$\sum \frac{(-1)^{n-1}}{n^{2}}$ is Convergent by the A.S.T

$a n=\frac{1}{n^{2}} \quad a_{1}=1 \quad a_{2}=\frac{1}{4} \quad a_{3}=\frac{1}{9} \quad a_{10}=\frac{1}{100}=0.01$

$a_{11}=\frac{1}{121}<0.01$$\Rightarrow$ from the estimation theory

$|S-S_{10} | \leq a_{11}=\frac{1}{121}<0.01$

We need to Consider $[n=10]$

The sum is approximated by $s_{10}$

$S_{10}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\frac{1}{64}+\frac{1}{81}+\frac{1}{100}$

$\approx 1.5497$