Inequality $\frac{\pi}{2}\leq\sum\limits_{n=0}^{\infty}\frac{1}{n^2+1}\leq 1+\frac{\pi}{2}$

Question

I need to prove that $$\frac\pi2 \le\sum_{n=0}^{\infty}\frac{1}{n^2+1}\le\frac\pi2+1.$$ Can anyone help me with that?

Answer 1

$\frac{1}{1+x^2}$ is a decreasing function on $\mathbb{R}^+$ and $$\int_{0}^{+\infty}\frac{dx}{x^2+1}=\frac{\pi}{2},$$ hence the given inequality is trivial by series-integral comparison.

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For a tighter inequality, we may notice that due to $\mathcal{L}^{-1}\left(\frac{1}{1+x^2}\right)=\sin(s)$ it follows that: $$ \sum_{n\geq 0}\frac{1}{1+n^2} = 1+\int_{0}^{+\infty}\frac{\sin(s)}{e^s-1}\,ds \stackrel{\text{Residues}}{=}\frac{1+\pi\coth\pi}{2} $$ and since $\coth(\pi)$ is pretty close to $1$, the given series is pretty close to $\frac{\pi+1}{2}$.

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A more elementary way is to exploit creative telescoping through: $$ \frac{1}{n^2+1} = \arctan\left(n+\tfrac{1}{2}\right)-\arctan\left(n-\tfrac{1}{2}\right) - \frac{C}{(n+1)(n+2)(n+3)},\quad 0\leq C \leq \tfrac{2}{3}$$ leading to $\sum_{n\geq 0}\frac{1}{n^2+1}\approx\frac{\pi}{2}+\arctan\tfrac{1}{2}$.

Answer 2

Actually We can compute $\sum_{n=0}^{\infty}\frac{1}{n^2+1}$ by Poisson summation formula.https://en.wikipedia.org/wiki/Poisson_summation_formula.

We compute its Fourier transform Let $f(x)=\frac{1}{1+x^2}$,then its fourier transform $\hat {f}(x)=\int_{-\infty}^{\infty}f(x)e^{-iwx}dx=\pi e^{-|w|}$.

then use summation formula.

$\sum_{n=-\infty}^{\infty}f(na)=\frac{1}{a}\hat{f}(\frac{2n\pi}{a})$,$a \neq 0$

Let $a=1$,$f(x)=\frac{1}{1+x^2}$.

$\sum_{n=-\infty}^{\infty}\frac{1}{n^2+1}=\pi\sum_{n=-\infty}^{\infty}e^{-2\pi|n|}=\pi coth\pi$.

$\sum_{n=-\infty}^{\infty}\frac{1}{n^2+1}=2\cdot\sum_{n=0}^{\infty}\frac{1}{n^2+1}-1$

Then your series converges to

$\frac{1+\pi coth\pi}{2}$

Then LHS obvious

As for RHS,We only need to prove

$coth\pi\le 1+\frac{1}{\pi}$

We only need to prove $\frac{e^{-2\pi}}{1-e^{-2\pi}}\le \frac{1}{2\pi}$

Let $g(t)=(t+1)e^{-t}-1$.

Show $g(t) \le 0$ when $t \ge 0$ yourself.Let $t=2\pi$.

Done.