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• Notes

Alternating Series:

$\sum_{n=1}^{\infty}(-1)^{n} b n$                    $n=1 \quad(-1)^1=-1$            $-b_{1}+b_{2}-b_{3}+b_4\cdots$

$\sum_{n=1}^{\infty}(-1)^{n-1} b n$                $n=1 \quad(-1)^{0}=1$                $b_{1}-b_{2}+b_{3}-b_{4} \dots$

Test for Alternating series: is convergent if:

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

Decreasing $\Rightarrow b_{n+1} \leq b_{n}$

Test the series $\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n}\right)$ for convergence.

A.S.T:

$\lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{1}{\infty}=0$

$f(x)=\frac{1}{x}$

$b_{n+1}=\frac{1}{n+1}<b n=\frac{1}{n}$

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

Test the series $\sum(-1)^{n} \frac{\ln (n)}{n}$ for convergence.

The sries is alternating $\Rightarrow$ Apply A.S.T:

$b n=\frac{\ln (n)}{n}$

$\lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\frac{\infty}{\infty}$   use L'Hopital Rule

$\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\lim _{n \rightarrow \infty} \frac{1/n}{1}=\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{1}{\infty}=0$

$f(x)=\frac{\ln (x)}{x} \Rightarrow f^{\prime}(x)=\frac{\frac{1}{x} \cdot x-(1)(\ln (x))}{x^{2}}=\frac{1-\ln (x)}{x^{2}}$

$x^{2} \Rightarrow$ is always positive$\ln (e)=1$    $, \ln (n) >1, n=3,4 \cdots$

$,\ln(n)< 1, n=1,2$

$1-\ln (x)<0$        $\Rightarrow$ for $x \geq 3$

$\{b n\}$ is decreasing

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

Test the series $\sum \frac{\cos (n \pi)}{n^{2}+1}$ for convergence.

Alternating Series

$\sum_{n=1}^{\infty}(-1)^{n} b n$

$b_{n}=\frac{1}{n^{2}+1}$

$\sum_{n=1}^{\infty} \cos (n \pi)=\cos (\pi)+\cos (2 \pi)+\cos (3 \pi) \ldots$

$=-1+1-1+1$

$=\sum_{n=1}^{\infty}(-1)^{n}$

$\sum \frac{\cos (n \pi)}{n^{2}+1}=\sum \frac{(-1)^{n}}{n^{2}+1}$ Alternating series

$b n=\frac{1}{n^{2}+1}$

Apply A.S.T  $\Rightarrow$

$\lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{1}{n^{2}+1}=\frac{1}{\infty^{2}}=\frac{1}{\infty}=0$

$f(x)=\frac{1}{x^{2}+1} \Rightarrow f^{1}(x)=\frac{-2 x(1)}{\left(x^{2}+1\right)^{2}}<0$

$\left\{b{n}\right\}$ is decreasing

$\sum \frac{\cos (n \pi)}{n^{2}+1}$ is Convergent by A.S.T