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Test the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}+1}\) for convergence or divergence.

Apply Integral test (I.T) 

\(f(n) \rightarrow a_{n}=\frac{1}{n^{2}+1} \rightarrow f(x)=\frac{1}{x^{2}+1}\)

(1) \(f(x) \text { is +ve valued on }[1, \infty)\)

(2) \(f(x) \text { is Continuons on }[1, \infty)\)

(3) \(f^{\prime}(x)=\frac{(0)\left(x^{2}+1\right)-(2 x)(1)}{\left(x^{2}+1\right)^{2}}=\frac{-2 x}{\left(x^{2}+1\right)^{2}}\)

\(f^{\prime}(x)=\frac{-2 x}{\left(x^{2}+1\right)^{2}}<0 \text { on }[1, \infty)\) decreasing

\(\int_{1}^{\infty} f(x) d x=\int_{1}^{\infty} \frac{1}{x^{2}+1} d x=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{x^{2}+1} d x\)

\(\int \frac{1}{x^{2}+1}=\tan ^{-1}(x)\)

\(\int_{1}^{\infty} f(x) d x=\lim _{t \rightarrow \infty} \tan ^{-1}\left.(x)\right|_{1} ^{t}\)

\(=\lim _{t \rightarrow \infty}\left[\tan ^{-1}(t)-\tan ^{-1}(1)\right]=\lim _{t \rightarrow \infty}\left[\tan ^{-1}(t)-\frac{\pi}{4}\right]\)

\(=\tan ^{-1}(\infty)-\frac{\pi}{4}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\)

\(\tan ^{-1}(\infty)=\frac{\pi}{2}\)

\(\int_{1}^{\infty} \frac{1}{x^{2}+1} d x\) is Convergent

\(\sum_{n=1}^{\infty} \frac{1}{n^{2}+1}\) is Convergent

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