real analysis - Computing $sum_{n = 1}^{+ infty} frac{1}{n^2 + 1}$ using Fourier series.

$f$ is a $2 \pi$ periodic function on $] - \pi,\pi[$ defined as :

$$f(x) = e^{-x}$$

Using Fourier series, compute the sums :

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

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How do I compute $\sum_{n = 1}^{+ \infty} \frac{1}{n^2 + 1}$ ?

I have computed the fourier coefficient and found that:

$$a_0 = \frac{2}{\pi} (1 - e^{- \pi})$$

$$a_n = \frac{2}{\pi(n^2 + 1)} (1 - (-1)^n . e^{- \pi} )$$

$$b_n = \frac{2n}{\pi(n^2 + 1)} (1 - (-1)^n . e^{- \pi} )$$

Using Dirichlet theorem and taking $x = 0$ to make the $b_n$ disappear, I get:

$$S_f (1) = \frac{2}{\pi} (1 - e^{- \pi}) + \sum_{n = 1}^{+ \infty} \frac{2}{\pi(n^2 + 1)} (1 - (-1)^n . e^{- \pi} ) . (-1)^n = 1$$

I do not see how to proceed to get the value of the sum?

The same problem for the second sum.

Answer

With the correct coefficients (see comments), you have for $x \in]-\pi,\pi[$:

$$e^{-x} = \frac{2}{\pi} \sinh(\pi) + 2 \sinh(\pi) \sum_{n=1}^{\infty}\frac{ (-1)^n (n\sin (nx) + \cos(nx)) }{\pi(n^2 + 1)}$$

so

$$e^{x} + e^{-x} = \frac{4}{\pi} \sinh(\pi) + 4 \sinh(\pi) \sum_{n=1}^{\infty}\frac{ (-1)^n \cos(nx) }{\pi(n^2 + 1)}$$

Now we can evaluate this for $x \to \pm\pi$. To be more precise (see the answer by @kvantour), the evaluation $x \to \pm\pi$ corresponds to the average of the two values at $\pm\pi$ which is exactly what the Fourier series converges to at this point of discontinuity.

$$\cosh (\pm \pi) = \frac{2}{\pi} \sinh(\pi)+ 2 \sinh(\pi) \sum_{n=1}^{\infty}\frac{1 }{\pi(n^2 + 1)}$$

and in turn

$$\sum_{n=1}^{\infty}\frac{1 }{(n^2 + 1)} = \frac{\pi \coth(\pi) -1}{2 }$$

Then the second question follows:
$$\sum_{- \infty}^{+ \infty} \frac{1}{n^2 + 1} = 1 + 2 \sum_{1}^{+ \infty} \frac{1}{n^2 + 1} = \pi \coth(\pi) \simeq 1.0037 \; \pi$$

Comment: for comparison,
$$\int_{- \infty}^{+ \infty} \frac{1}{x^2 + 1} {\rm{dx}} = \pi$$