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• Notes

Test the convergence of $\sum_{n=1}^{\infty}\left[\tan ^{-1}(n+1)-\tan ^{-1}(n)\right]$ and if it is convergent, Find its sum

$a_{n}=\tan ^{-1}(n+1)-\tan ^{-1}(n)$

$S n=a_{1}+a_{2}+a_{3}+\dots-a n$

$=\left[\tan ^{-1}(2)-\tan ^{-1}(1)\right]+\left[\tan ^{-1}(3)-\tan (2)\right]$

$+\cdots \cdot\left[\tan^{-1} (n+1)-\tan ^{-1}(n)\right]$

$\begin{array}{l}{a_{1}=\tan ^{-1}(2)-\tan ^{-1}(1)} \\ {a_{2}=\tan ^{-1}(3)-\tan ^{-1}(2)} \\ {a_{3}=\tan ^{-1}(4)-\tan ^{-1}(3)} \\ {a_{n}=\tan ^{-1}(n+1)-\tan ^{-1}(n)}\end{array}$

the only remaining terms are:

$S_{n}=\tan ^{-1}(n+1)-\tan ^{-1}(1)$

$=\tan ^{-1}(n+1)-\frac{\pi}{4}$

$\lim _{n \rightarrow \infty} {sn}=\lim _{n \rightarrow \infty}\left[\tan ^{-1}(n+1)-\frac{\pi}{a}\right]$

$=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}= {sum}$

$\sum_{n=1}^{\infty}\left[\tan ^{-1}(n+1)-\tan ^{-1}(n)\right]$ is convergent

and its sum $=\frac{\pi}{4}$

Test the convergence of $\sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right)$ and if it is convergent, Find its sum

$\ln \left(\frac{a}{b}\right)=\ln (a)-\ln (b)$

$a n=\ln \left(\frac{n+1}{n}\right)=\ln (n+1)-\ln (n)$

$\begin{array}{l}{a_{1}=\ln (2)-\ln (1)} \\ {a_{2}=\ln (3)-\ln (2)} \\ {a_{3}=\ln (4)-\ln (3)} \\ {\vdots} \\ {a_{n}=\ln (n+1)-\ln (n)}\end{array}$

$S n=a_{1}+a_{2}+a_{3}+a_{4} \dots+a_{n}=\ln (n+1)$

$\lim _{n \rightarrow \infty} s{n}=\lim _{n \rightarrow \infty} \ln (n+1)=\ln (\infty+1)=\ln (\infty)=\infty$

$\sum_{n=1}^{\infty} \ln \left(\frac{(n+1 )}{n}\right)$ is Divergent and has no sum

Find the sequence of partial sums, then test the convergence for the following series $\sum_{n=1}^{\infty} \frac{2}{4 n^{2}-1}$

$a n=\frac{2}{4 n^{2}-1}=\frac{2}{(2 n+1)(2 n-1)}$

$=\frac{A}{2 n-1}+\frac{B}{2 n+1}$

$A=\frac{2}{(2 n+1)(2 n-1)} \cdot\left.(2 n-1)\right|_{n=\frac{1}{2}}$

$=\left.\frac{2}{2 n+1}\right|_{n=\frac{1}{2}}=\frac{2}{2 \cdot \frac{1}{2}+1}=\frac{2}{2}=1$

$B=\frac{2}{(2 n+1)(2 n-1)} \cdot\left.(2 n+1)\right|_{n=-\frac{1}{2}}$

$=\left.\frac{2}{2 n-1}\right|_{n=-\frac{1}{2}} =\frac{2}{2-\frac{1}{2}-1}=\frac{2}{-2}=-1$

$a n=\frac{1}{2 n-1}+\frac{-1}{2 n+1}$

$S_{n}=a_{1}+a_{2}+a_{3}+\dots a n$

$a n=\frac{1}{2 n-1}-\frac{1}{2 n+1}$

$S n=a_{1}+a_{2}+a_{3}+\dots a_{n}$

$\begin{array}{l}{a _1=\frac{1}{1}-\frac{1}{3}} \\ {a_{2}=\frac{1}{3}-\frac{1}{5}} \\ {a_{n}=\frac{1}{7}-\frac{1}{9}} \\ {\vdots} \\ {a n=\frac{1}{2 n-1}-\frac{1}{2 n+1}}\end{array}$

$\Longrightarrow\{s {n}\}=\left\{1-\frac{1}{2 n+1}\right\}$

$\lim _{n \rightarrow \infty} {sn}=\lim _{n \rightarrow \infty}\left(1-\frac{1}{2 n+1}\right)=1$

${sum}=1$

$\sum_{n=1}^{\infty} \frac{2}{4 n^{2}-1}$ is convergent and its sum = 1