Sum to Infinity of 2x^2n over n by 2n Choose n

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Theorem

For $\cmod x < 1$:

$\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} } = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}$


Proof

By Gregory Series:

$\ds \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$


Let $t = \dfrac x {\sqrt {1 - x^2} }$.

Let $y = \arcsin x$.

Then:

\(\ds t\) \(=\) \(\ds \frac {\sin y} {\sqrt {1 - \sin^2 y} }\)
\(\ds \) \(=\) \(\ds \frac {\sin y} {\cos y}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \tan y\)

Hence $\arctan t = \arcsin x$.


We have:

\(\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} }\) \(=\) \(\ds 2 t \arctan t\)
\(\ds \) \(=\) \(\ds 2 t \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}\) Gregory Series
\(\ds \) \(=\) \(\ds 2 t \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m - 1} } {2 m - 1}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m} } {2 m - 1}\)
\(\ds \) \(=\) \(\ds 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {\paren {2 m - 1} \paren {1 - x^2}^m}\)
\(\ds \) \(=\) \(\ds 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {2 m - 1} \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}\) Binomial Theorem for Negative Index and Negative Parameter

It remains to show the the coefficient of $x^{2 n}$ on the right hand side is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is:

$\ds 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$

The left hand side above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\ds \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$.


We have:

\(\ds \) \(\) \(\ds 2 n \dbinom {2 n} n \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {n - 1} {r - 1}\)
\(\ds \) \(=\) \(\ds 2 n \dbinom {2 n} n \sum_{r \mathop = 0}^{n - 1} \frac {\paren {-1}^r} {2 r + 1} \dbinom {n - 1} r\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds 2 n \dbinom {2 n} n \int_0^1 \sum_{r \mathop = 0}^{n - 1} \paren {-1}^r \dbinom {n - 1} r y^{2 r} \d y\)
\(\ds \) \(=\) \(\ds 2 n \dbinom {2 n} n \int_0^1 \paren {1 - y^2}^{n - 1} \d y\) Binomial Theorem
\(\ds \) \(=\) \(\ds 2 n \dbinom {2 n} n \int_{\frac \pi 2}^0 \sin^{2 n - 2} \theta \, \frac {\d y} {\d \theta} \d \theta\) by substitution of $y = \cos \theta$
\(\ds \) \(=\) \(\ds 2 n \dbinom {2 n} n \int_0^{\frac \pi 2} \sin^{2 n - 1} \theta \, \d \theta\)
\(\ds \) \(=\) \(\ds 2 n \dbinom {2 n} n \frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!}\) Definite Integral from 0 to Half Pi of Odd Power of Sine x
\(\ds \) \(=\) \(\ds 2 n \paren {\frac {\paren {2 n}!} {n! \, n!} } \paren {\frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!} }\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds 2 n \paren {\frac {2 n} {n^2} } \paren {2^{2 n - 2} }\)
\(\ds \) \(=\) \(\ds 2^{2 n}\)

Hence the result.

$\blacksquare$


Sources

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