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
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $4$.