Evaluate the series:
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4^n n^2} \binom{2n}{n}$$
Solution
Standard technics of this series get stuck at the evaluation of polylogarithm value. Therefore, we are attacking the series in a non direct manner.
Let us consider the function
$$f(a)=\int_{0}^{\pi/2}\log \left ( 1+a \cos^2 \theta \right )\, {\rm d}\theta, \;\; a\geq 0$$
Hence,
$$ \begin{align*} f'(a) &=\int_{0}^{\pi/2}\frac{\cos^2 \theta}{1+a\cos^2 \theta}\, {\rm d}\theta \\ &=\int_{0}^{\pi/2} \frac{{\rm d}\theta}{\frac{1}{\cos^2 \theta}+a} \\ &=\int_{0}^{\pi/2}\frac{{\rm d}\theta}{\tan^2 \theta +1 +a} \\ &=\frac{\pi}{2}\cdot \frac{1}{a+\sqrt{a+1}+1} \end{align*}$$
Integrating back (and at the same time taking care of the constant) we have that:
\begin{equation} f(a)= \pi \log \left ( \sqrt{a+1}+1 \right ) - \pi \log 2 \end{equation}
Setting $a=1$ back at $(1)$ we have that:
\begin{align*} f(1) &=\pi \log \left ( \sqrt{2}+1 \right ) -\pi \log 2 \\ &=\int_{0}^{\pi/2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1} \cos^{2n}\theta}{n} \, {\rm d}\theta \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{0}^{\pi/2} \cos^{2n} \theta \, {\rm d}\theta\\ &= \frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\Gamma \left ( n+\frac{1}{2} \right )}{n!}\\ &=\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\frac{\left ( 2n \right )!}{4^n (n!)^2} \\ &= \frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^n n^2}\binom{2n}{n} \end{align*}
We used Wallis formula for the evaluation of $\displaystyle \int_0^{\pi/2} \cos^{2n} x \, {\rm d}x$ and the known formula for $\Gamma \left(n +\frac{1}{2} \right)$. Thus:
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4^n n^2} \binom{2n}{n}=2\log \left(\sqrt{2}+1\right)-2\log 2$$
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4^n n^2} \binom{2n}{n}$$
Solution
Standard technics of this series get stuck at the evaluation of polylogarithm value. Therefore, we are attacking the series in a non direct manner.
Let us consider the function
$$f(a)=\int_{0}^{\pi/2}\log \left ( 1+a \cos^2 \theta \right )\, {\rm d}\theta, \;\; a\geq 0$$
Hence,
$$ \begin{align*} f'(a) &=\int_{0}^{\pi/2}\frac{\cos^2 \theta}{1+a\cos^2 \theta}\, {\rm d}\theta \\ &=\int_{0}^{\pi/2} \frac{{\rm d}\theta}{\frac{1}{\cos^2 \theta}+a} \\ &=\int_{0}^{\pi/2}\frac{{\rm d}\theta}{\tan^2 \theta +1 +a} \\ &=\frac{\pi}{2}\cdot \frac{1}{a+\sqrt{a+1}+1} \end{align*}$$
Integrating back (and at the same time taking care of the constant) we have that:
\begin{equation} f(a)= \pi \log \left ( \sqrt{a+1}+1 \right ) - \pi \log 2 \end{equation}
Setting $a=1$ back at $(1)$ we have that:
\begin{align*} f(1) &=\pi \log \left ( \sqrt{2}+1 \right ) -\pi \log 2 \\ &=\int_{0}^{\pi/2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1} \cos^{2n}\theta}{n} \, {\rm d}\theta \\ &= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{0}^{\pi/2} \cos^{2n} \theta \, {\rm d}\theta\\ &= \frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\frac{\Gamma \left ( n+\frac{1}{2} \right )}{n!}\\ &=\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\frac{\left ( 2n \right )!}{4^n (n!)^2} \\ &= \frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^n n^2}\binom{2n}{n} \end{align*}
We used Wallis formula for the evaluation of $\displaystyle \int_0^{\pi/2} \cos^{2n} x \, {\rm d}x$ and the known formula for $\Gamma \left(n +\frac{1}{2} \right)$. Thus:
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4^n n^2} \binom{2n}{n}=2\log \left(\sqrt{2}+1\right)-2\log 2$$
However it is known that:
ReplyDelete\begin{equation} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\binom{2n}{n}x^n = \log^2 \left ( \frac{1+\sqrt{1+4x}}{2} \right ) - 2\operatorname{Li}_2 \left ( \frac{1- \sqrt{1+4x}}{2} \right ) \end{equation}
This can be extracted by setting $x \mapsto -x$ in the known generating series
$$\sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{x^n}{n} = -2\log \left(\frac{1+\sqrt{1-4x}}{2}\right)$$
Now setting $x=\frac{1}{4}$ back at $(2)$ we get that:
$$\operatorname{Li}_2 \left ( \frac{1-\sqrt{2}}{2} \right ) = \frac{1}{2}\log^2 \left ( \frac{1+\sqrt{2}}{2} \right ) - \log \left ( \sqrt{2}+1 \right )+\log 2$$
I like the first method, T. :)
ReplyDeleteThanks C... I also like that method! Playing around, well, I guess, we can produce more polylog values... what do you think?
Delete