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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

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