# Sum of $\dfrac{1}{n^2+1}$ using Parseval's theorem

I know this question has been widely answered here, but without using Fourier analysis. Also there is a video referring to this trick but I want to use a different Fourier series.

First of Parseval's Theorem states: $$\displaystyle{\dfrac{1}{\pi}\int_{-\pi}^{\pi}[f(x)]^2 = [a_0]^2+\sum_{n=1}^{\infty}[a_n]^2+[b_n]^2}$$.

I calculate $$a_0$$ like $$\frac{1}{2\,\pi}\,\int_{-\pi}^{\pi}f(x)\,\mathrm{dx}$$ instead of $$\frac{1}{\pi}\,\int_{-\pi}^{\pi}f(x)\,\mathrm{dx}$$, so no need for $$\frac{1}{2}\,a_0$$

Here I'd like to involve the series of $$\cosh(x)$$ whose partition I have been asked for in a prior question.

It should be: $$\displaystyle{\cosh(x) = \underbrace{\dfrac{\sinh(\pi)}{\pi}}_{a_0}+\sum_{n = 1}^{\infty}\underbrace{\color{red}{2}\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n\,\cos(n\,x)}_{a_n}}$$

Plugging those terms into the original theorem:

\begin{align} &\dfrac{1}{\pi}\int_{-\pi}^{\pi}\cosh^2(\pi) = \left(\dfrac{\sinh(\pi)}{\pi}\right)^2+\sum_{n=1}^{\infty}\left(\color{red}{2}\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n\right)^2 \\\\ & 1+\dfrac{\sinh(2\,\pi)}{2\,\pi}= \dfrac{\sinh^2(\pi)}{\pi^2}\,\left[1+\sum_{n=1}^{\infty}\color{red}{4}\,\left(\dfrac{1}{1+n^2}\right)^2\right]\\\\ &\sqrt{\left(\dfrac{\pi^2}{\sinh^2(\pi)}+\dfrac{\sinh(2\,\pi)\,\pi}{2\,\sinh^2(\pi)}-1\right)\,\color{red}{\dfrac{1}{4}}} = \sum_{n=1}^{\infty}\dfrac{1}{1+n^2} \quad ? \end{align}

Actually the value is coming close to the approximated sum, but it's not exactly the same result...

Edit

I fixed the Fourier Series highlighting the missing term in red. On the other hand I eradicated some factors, not sure about making it worse. It's still differing from the approximation.

approximation $$\approx 1,0767$$

Therefore it works exactly like suggested by Stefan Lafon in the remarks: setting $$x=\pi$$

\begin{align} &\cosh(\pi) = \dfrac{\sinh(\pi)}{\pi}+\displaystyle{\sum_{n=1}^{\infty}2\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}}\\\\ &\Rightarrow \quad \dfrac{\cosh(\pi)\,\pi}{2\,\sinh(\pi)}-\dfrac{1}{2} =\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{1+n^2}} \end{align}

It just remains a mystery why the method on top fails

• I don't think you should be using Parseval. Rather, use the Fourier series expansion that you have for $\cosh(x)$, and evaluate at $x=\pi$. Jul 25, 2021 at 21:32

Note that when you did Parseval, you had (ignoring other pieces in the expression)

$$\sum_{n=1}^{\infty} \frac{1}{(1+n^2)^{\color{red}2}}.$$

Note the exponent in red. This is not $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{1+n^2}$$ and is the source of the issues you are having.

Per Stefan Lafon (for posterity and the purposes of having an answer to this question): simply evaluate your Fourier series at $$x=\pi$$ to get the correct expression which is

$$\frac{\pi}{2} \operatorname{coth}(\pi) - \frac{1}{2}.$$

• I thought taking the square root would have fixed the issue but apparently it's changing the sum. Anyway the other method has worked.
– Leon
Jul 27, 2021 at 19:36
• Recall that $(a^2 + b^2)^{1/2} \neq a + b$ in general. This is just an extension of that fact. Jul 27, 2021 at 19:43