Sum of the series $\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$ How do I find the sum of the following infinite series:
$$\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$$
I think the sum can be converted to definite integral and calculated but I don't know how to proceed from there.
 A: The $n$-th term in the series is $\dfrac{2 \cdot 6 \cdots (2n-2)}{5 \cdot 10 \cdots 5n} = \dfrac{1}{n+1}\dbinom{2n}{n}\dfrac{1}{5^{n+1}} = \dfrac{C_n}{5^{n+1}}$, where $C_n = \dfrac{1}{n+1}\dbinom{2n}{n}$ is the $n$-th Catalan number. Thus, the sum is $\displaystyle\sum_{n = 1}^{\infty}\dfrac{C_n}{5^{n+1}}$.
The generating function of the Catalan numbers is $c(x) = \displaystyle\sum_{n = 0}^{\infty}C_nx^n = \dfrac{1-\sqrt{1-4x}}{2x}$. 
Multiply by $x$ and subtract $x$ to get $xc(x)-x = \displaystyle\sum_{n = 1}^{\infty}C_nx^{n+1} = \dfrac{1-\sqrt{1-4x}}{2} - x$
Then, plug in $x = \dfrac{1}{5}$ to get $\displaystyle\sum_{n = 1}^{\infty}\dfrac{C_n}{5^{n+1}} = \dfrac{3-\sqrt{5}}{10}$.
A: Using generalized Binomial Theorem, $$(1+z)^n=1+nz+\frac{n(n-1)}{2!}z^2+\frac{n(n-1)(n-2)}{3!}z^3+\cdots$$  for $|z|<1$
$\displaystyle S=\sum_{r=1}^\infty\dfrac{2\cdot6\cdots(4r-2)}{5\cdot10\cdots5(r+1)}$
Observe that there are $r$ terms in the numerator unlike the denominator which has $r+1$ terms
So, we write multiplying the numerator with the previous term of $2$ of the Arithmetic Series $2,6,10,\cdots,$
$\displaystyle -2S=\sum_{r=1}^\infty\dfrac{(-2)2\cdot6 \cdots(4r-2)}{5\cdot10\cdots5(r+1)}$
Dividing each multiplicand of the numerator by the common difference of $-2,2,6,\cdots$ and taking out $5$ from each multiplicand of the denominator, 
$\displaystyle -2S=\sum_{r=1}^\infty\frac{4^{r+1}}{5^{r+1}}\cdot\dfrac{-\dfrac24\cdot\dfrac24\cdot\dfrac64\cdots\dfrac{4r-2}4}{(r+1)!}$
Multiplying each multiplicand  of the numerator by $-1$ as the formula needs $r(r-1)(r-2)\cdots,$
$\displaystyle -2S=\sum_{r=1}^\infty\left(\frac45\right)^{r+1}\cdot\dfrac{\frac12\left(\frac12-1\right)\cdot\left(\frac12-2\right)\cdots\left(\frac12-r\right)}{(r+1)!}\cdot(-1)^{r+1}$
Observe that here the denominator and the exponent of $-\dfrac45$ of the first term $=2!,$ and compare with the generalized Binomial expansion
$\displaystyle\implies -2S+1+\frac12\left(-\frac45\right)$
$\displaystyle=1+\frac12\left(-\frac45\right)+\sum_{r=1}^\infty\dfrac{\frac12\left(\frac12-1\right)\cdot\left(\frac12-2\right)\cdots\left(\frac12-r\right)}{(r+1)!}\left(-\frac45\right)^{r+1}$
$\displaystyle\implies -2S+1+\frac12\left(-\frac45\right)=\left(1-\frac45\right)^{\dfrac12}$
Hope things are manageable from here
A: Less simple and less elegant than JimmyK4542's answer : for each term of the summation, the numerator is $$N_n=\prod_{i=1}^n (4i-2)=\frac{4^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}$$ and the denominator is $$D_n=\prod_{i=1}^{n+1} (5i)=5^{n+1} \Gamma (n+2)$$and we have to compute $$S_n=\sum_{i=1}^n \frac{N_i}{D_i}=\sum_{i=1}^n \frac{4^n  \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }5^{n+1}\Gamma (n+2)}$$ from which can be derived the result already given for $S_{\infty}$
