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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 \(s_{10}\)

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

\( \approx 1.5497 \)

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