$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $

Question

Prove that:

$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $

What I've tried:

I solved the improper integral: $\int_0^\infty \frac{1}{1+x^2} = \lim_{b\to \infty} \arctan b -\arctan 0 = \frac{\pi}2 $.

Now, everything in the sum (and in the integral) is positive so the sum must be lower than the integral (it has more members), so the claim is wrong (originally it was a "Prove/Disprove" problem...). However, if we only add the first 3 members we get 1.7 which is bigger than $\frac{\pi}2 $. Also, I checked with wolfram and the sum is somewhere around 2 so the claim is true. I have no idea how to prove it.

Someone suggested I would read Basel's problem ,in which the integral is also smaller than the sum (starting from 1) but it led me nowhere and I don't think that this kind of complex solution is needed here.

Answer 1

The key result to use is that if $f(x)$ is a continuous strictly decreasing function, then $f(n+1) < \int_{n}^{n+1} f(x)\,dx < f(n)$.

If we take the sum over all values of $n$ starting at $n=0$ using the right inequality, we get that $$\frac{\pi}{2} = \int_{0}^\infty \frac{dx}{1+x^2} < \sum_{n=0}^\infty \frac{1}{1+n^2},$$

and if we sum over all values of $n$ starting at $n=1$ and use the left inequality, we get $$ \sum_{n=2}^\infty \frac{1}{1+n^2} < \int_{1}^\infty \frac{dx}{1+x^2} = \frac{\pi}{4}.$$
We then get the desired inequality as follows: $$ \sum_{n=0}^\infty \frac{1}{1+n^2} = 1 + \frac{1}{2} + \sum_{n=2}^\infty \frac{1}{1+n^2} < \frac{3}{2} + \frac{\pi}{4} < \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}.$$ (This elaboration is necessary since $\frac{1}{1+x^2}$ is increasing on $[-1,0]$, as noted in comments.)

Answer 2

An advanced approach using zeta function values $\zeta(2)$ and $\zeta(4)$ to get even better bounds.

For $n>1$, $$\frac{1}{1+n^2}=\frac{1}{n^2}\frac{1}{1+\frac{1}{n^2}} = \frac1{n^2}-\frac{1}{n^4}+\frac{1}{n^6}-\frac{1}{n^8}+...$$

So $$\frac{1}{n^2}-\frac{1}{n^4}<\frac{1}{n^2+1}<\frac{1}{n^2}$$

So:

$$\frac{3}{2}+\left(\zeta(2)-\zeta(4)-\left(\frac{1}{2^2}-\frac{1}{2^4}\right)\right)=1+\frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n^2} -\frac{1}{n^4}\right)<\sum_{n=0}^{\infty}\frac{1}{n^2+1} < \frac{1}{2} + \zeta(2)$$

Now, $$\frac{1}{2}+\zeta(2)=\frac 12+\frac{\pi^2}{6} <\frac{3\pi}{4}$$

And:

$$\frac{21}{16}+\zeta(2)-\zeta(4) = \frac{21}{16} + \frac{\pi^2}{6} - \frac{\pi^4}{90}>\frac{\pi}{2}$$

Answer 3

Of course, one could be very silly and note that the sum in question has a well-known exact value:

$$\sum_{n=0}^{\infty} \frac1{1+n^2} = \frac{\pi}{2} \operatorname{coth}{\pi} + \frac12 $$

Now

$$\operatorname{coth}{\pi} = \frac{1+e^{-2 \pi}}{1-e^{-2 \pi}} \approx 1+2 e^{-2 \pi} \approx 1$$

It is not hard to show that the inequality is true from this.

Answer 4

Maybe it's interesting to see why we have the closed form $$\sum_{n\geq0}\frac{1}{1+n^{2}}=\frac{\pi}{2}\coth\left(\pi\right)+\frac{1}{2} $$ and a way is to use this well known result from complex analysis: $$\sum_{n=-\infty}^{\infty}f\left(n\right)=-\pi\sum_{k=1}^{m}\textrm{Res}\left(f\left(z\right)\cot\left(\pi z\right),a_{k}\right) $$ where $a_{k} $ are all poles of $f\left(z\right) $. In this case the poles are $\pm i $, so, observing that also holds $$ \sum_{n=-\infty}^{\infty}\frac{1}{1+n^{2}}=2\sum_{n\geq0}\frac{1}{1+n^{2}}-1 $$ we have $$ \sum_{n\geq0}\frac{1}{1+n^{2}}=\frac{\pi\coth\left(\pi\right)}{2}+\frac{1}{2} $$ and it's easily to see that, as Ron Gordon showed, this number is between $\pi/2 $ and $3\pi/4 $.