Sum of Reciprocals of Squares plus 1

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Theorem

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 + 1} = \frac 1 2 \paren {\pi \coth \pi - 1}$


Proof

Let $\map f x$ be the real function defined on $\hointl {-\pi} \pi$ as:

$\map f x = e^x$


From Fourier Series: $e^x$ over $\openint {-\pi} \pi$, we have:

$(1): \quad \ds \map f x \sim \map S x = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} } }$


Let $x = \pi$.

By Fourier's Theorem, we have:

\(\ds \map S \pi\) \(=\) \(\ds \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}\)
\(\ds \) \(=\) \(\ds \dfrac {e^\pi + e^{-\pi} } 2\)
\(\ds \) \(=\) \(\ds \cosh \pi\) Definition of Hyperbolic Cosine


Thus setting $x = \pi$ in $(1)$, we have:

\(\ds \cosh \pi\) \(=\) \(\ds \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos \pi} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} } }\)
\(\ds \leadsto \ \ \) \(\ds \frac {\pi \cosh \pi} {\sinh \pi}\) \(=\) \(\ds 1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\paren {-1}^n} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} }\) Cosine of Multiple of Pi and multiplying both sides by $\dfrac \pi {\sinh \pi}$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\pi \coth \pi - 1} 2\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {1 + n^2}\) Definition 2 of Hyperbolic Cotangent, Sine of Multiple of Pi and simplifying

$\blacksquare$


Sources