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.

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 $\displaystyle\sum_{n = 1}^{\infty}\dfrac{C_n}{5^{n+1}} = \dfrac{3-\sqrt{5}}{10}$.
