Sum to Infinity of Reciprocal of n^2 by 2n Choose n

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Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 \dbinom {2 n} n}\) \(=\) \(\ds \frac {\pi^2} {18}\)
\(\ds \) \(\approx\) \(\ds 0 \cdotp 54831 \, 13556 \ldots\)

This sequence is A086463 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

By Sum to Infinity of 2x^2n over n by 2n Choose n, for $0 < \size x < 1$:

\(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}\) \(=\) \(\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} }\)
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n - 1} } {n \dbinom {2 n} n}\) \(=\) \(\ds \frac {\arcsin x} {\sqrt {1 - x^2} }\)
\(\ds \leadsto \ \ \) \(\ds \int \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n - 1} } {n \dbinom {2 n} n} \d x\) \(=\) \(\ds \int \frac {\arcsin x} {\sqrt {1 - x^2} } \d x\) integrating both sides with respect to $x$
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac {2^{2 n - 1} x^{2 n} } {2 n^2 \dbinom {2 n} n}\) \(=\) \(\ds \int \arcsin x \d \paren {\arcsin x}\) Derivative of Arcsine Function
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\arcsin x}^2 + C\) Integration by Substitution

Substituting $x = 0$ gives:

$0 = 0 + C$

so $C = 0$.

Hence:

\(\ds \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n^2 \dbinom {2 n} n}\) \(=\) \(\ds 2 \paren {\arcsin x}^2\)
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 \dbinom {2 n} n}\) \(=\) \(\ds 2 \paren {\arcsin \frac 1 2}^2\) substituting $x = \frac 1 2$
\(\ds \) \(=\) \(\ds 2 \paren {\frac \pi 6}^2\) Sine of $30^\circ$
\(\ds \) \(=\) \(\ds \frac {\pi^2} {18}\)

$\blacksquare$


Also see


Sources

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