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Now I am not by any means skilled in this level of calculus, but I don't understand something. When I use Fourier series and calculate the coefficients by using $f(x)=e^x$, and plug in $x=\pi$, I get something completely different for $\sum_{n=1}^{\infty}\frac{1}{n^2+1}$. In fact, I obtain $\frac{\pi{e^{\pi}}}{2\sinh\pi}-\frac{1}{2}$. Can someone tell me why a different answer arises from this rather than using the complex fourier series with $c_n$? Perhaps there is something I am missing with Parseval's theorem? (P.S. I just learned this, so jargon might fly over my head. I am also new to forums completely).

Jyrki Lahtonen
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TheBean101007
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    Exactly what kind of a Fourier series did you use to sum this series? – Jyrki Lahtonen Apr 26 '21 at 04:49
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    Without seeing your calculation we cannot comment if there is an error. – Ross Millikan Apr 26 '21 at 05:02
  • I used the series that looks like this, (not sure of the technical term): $f(x)=a_o+\sum_{n=1}^{\infty}a_ncos(nx)+b_nsin(nx)$. When I calculated the coefficients for $a_o$, $a_n$, and $b_n$, I got $\frac{2sinh\pi}{2\pi}$, $\frac{(-1)^n2sinh\pi}{(n^2+1)\pi}$ , and $\frac{(-1)^{n+1}2nsinh\pi}{\pi(n^2+1)}$ respectively. When I plug in x=pi, the sin terms cancel out to zero, and the cos(npi) becomes (-1)^n, and that squared becomes simply 1 on the numerator which leaves me my sum and e^pi on the other side, but that seeks to be incorrect. I looked up the answer, and mine was wrong. Why? – TheBean101007 Apr 26 '21 at 21:33

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