1. ## infinite series

Prove that

$\sum_{k=0}^{\infty} \frac{ (-1)^k}{ \binom{2k}{k} }$

$= \frac{4}{5} [ 1 - \frac{1}{\sqrt{5}} \ln( \frac{ 1 + \sqrt{5}}{2} ) ]$

and

$\sum_{k=0}^{\infty} \frac{1}{ \binom{2k}{k} }$

$= \frac{ 2( \pi + 6\sqrt{3})}{9\sqrt{3}}$

and finally what is $\sum_{k=0}^{\infty} \frac{1}{ \binom{4k}{2k} }$ ??

2. Originally Posted by simplependulum
Prove that

$\sum_{k=0}^{\infty} \frac{ (-1)^k}{ \binom{2k}{k} }$

$= \frac{4}{5} [ 1 - \frac{1}{\sqrt{5}} \ln( \frac{ 1 + \sqrt{5}}{2} ) ]$

and

$\sum_{k=0}^{\infty} \frac{1}{ \binom{2k}{k} }$

$= \frac{ 2( \pi + 6\sqrt{3})}{9\sqrt{3}}$

and finally what is $\sum_{k=0}^{\infty} \frac{1}{ \binom{4k}{2k} }$ ??
let $f(\alpha)=1 + \sum_{k \geq 1} \frac{\alpha^k}{\binom{2k}{k}}, \ \ 0 \neq |\alpha| \leq 1.$ your question is to find $f(-1), \ f(1),$ and $\frac{f(1)+f(-1)}{2}.$ we have: $\frac{1}{\binom{2k}{k}}=\frac{(k!)^2}{(2k)!}=\frac {k \Gamma(k) \Gamma(k+1)}{\Gamma(2k+1)}=kB(k,k+1)=k \int_0^1 x^{k-1}(1-x)^k \ dx.$

hence: $f(\alpha)=1 + \sum_{k \geq 1}k \alpha^k \int_0^1 x^{k-1}(1-x)^k \ dx= 1 + \int_0^1 \frac{1}{x} \sum_{k \geq 1} k(\alpha x - \alpha x^2)^k \ dx.$ therefore: $f(\alpha)= 1 + \int_0^1 \frac{\alpha(1-x)}{(\alpha x^2 - \alpha x + 1)^2} \ dx=1 + \frac{1}{\alpha} \int_0^1 \frac{1-x}{(x^2 - x + \frac{1}{\alpha})^2} \ dx.$

finally substituting $x \to x - \frac{1}{2}$ gives us: $f(\alpha) = 1 + \frac{1}{\alpha} \int_0^{\frac{1}{2}} \frac{dx}{(x^2 + \beta)^2},$ where $\beta = \frac{1}{\alpha} - \frac{1}{4}.$

Example: let's find $f(1)$ : we have $f(1)=1 + \int_0^{\frac{1}{2}} \frac{dx}{(x^2 + \frac{3}{4})^2}.$ let $x = \frac{\sqrt{3}}{2} \tan t.$ then $\int_0^{\frac{1}{2}} \frac{dx}{(x^2 + \frac{3}{4})^2}=\frac{8 \sqrt{3}}{9} \int_0^{\frac{\pi}{6}} \cos^2 t \ dt=\frac{1}{3} + \frac{2 \sqrt{3} \pi}{27}.$ therefore: $f(1)=\frac{4}{3} + \frac{2 \sqrt{3} \pi}{27}.$