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.

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
