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How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ ?

My Attempt : I was using Integral Test of a Series. I got $\int_0^\infty \frac{1}{1+x^2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le \frac{1}{1+0^2} + \int_0^\infty \frac{1}{1+x^2}$ which gives $ \frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le 1+ \frac{\pi}{2}$.

Can anyone please help me by giving any hint ?

anonymous
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2 Answers2

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We shall give a stronger bound. Note that $$ 2=1+\sum^\infty_{n=1}\frac{1}{n^2+n}<\sum^\infty_{n=0}\frac{1}{n^2+1}<1+\frac{1}{2}+\sum^\infty_{n=2}\frac{1}{n^2}=\frac{1}{2}+\frac{\pi^2}{6}$$ which is a stronger bound.

Zhang
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This can be computed using the result of this answer: $$ \begin{align} \sum_{n=0}^\infty\frac1{n^2+1} &=-\frac1{2i}\sum_{n=0}^\infty\left(\frac1{i-n}+\frac1{i+n}\right)\\ &=\frac12-\frac1{2i}\sum_{n\in\mathbb{Z}}\frac1{i+n}\\ &=\frac12-\frac1{2i}\pi\cot(\pi i)\\[3pt] &=\frac12+\frac\pi2\coth(\pi)\\[9pt] &\approx2.076674 \end{align} $$ We can get bounds tighter than $\frac\pi2$ and $\frac{3\pi}4$ with telescoping series: $$ \begin{align} \sum_{n=0}^\infty\frac1{n^2+1} &\ge1+\sum_{n=1}^\infty\frac1{n^2+n}\\ &=1+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)\\[6pt] &=2 \end{align} $$ and, although this does not give as tight an upper bound as in Zhang's answer, it only uses telescoping series $$ \begin{align} \sum_{n=0}^\infty\frac1{n^2+1} &\le1+\frac12+\sum_{n=2}^\infty\frac1{n^2-\frac14}\\ &=1+\frac12+\sum_{n=2}^\infty\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\[3pt] &=\frac{13}6 \end{align} $$ Thus, $$ 2\le\sum_{n=0}^\infty\frac1{n^2+1}\le\frac{13}6 $$

robjohn
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