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• Notes

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