The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we reached at this point
$$= 1/20 - [1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
Please help someone in solving this question. I am very grateful to you.

This second sum can be written as $$\sum_{n=2}^{\infty}\frac{1}{n}\left(\frac{1}{5}\right)^n = \frac{1}{25}\sum_{n=0}^{\infty}\frac{1}{n+2}\left(\frac{1}{5}\right)^n$$
This is the Lerch Transendent, so $$=\frac{1}{25}\phi\left(\frac{1}{5}, 1,2\right)$$
By the identities $\phi(z,s,a) = \frac{1}{z}\left(\phi(z,s,a-1) - \frac{1}{((a-1)^2)^{s/2}}\right)$, $\frac{1}{z}\mathrm{Li}_n(z)=\phi(z,n,1)$ and $\mathrm{Li}_1(z)=-\ln(1-z)$:
$$\begin{align} &=\frac{1}{25}\cdot5\left(\phi\left(\frac{1}{5},1,1\right)-1\right) \\ &= \frac{1}{5}\left(5\,\mathrm{Li}\left(\frac{1}{5}\right)-1\right) \\ &= -\ln\left(\frac{4}{5}\right)-\frac{1}{5}\end{align}$$
Therefore, the final answer to your entire sum is $$\begin{align} &= \frac{1}{20} - \left(-\ln\left(\frac{4}{5}\right)-\frac{1}{5}\right)\\ &= \frac{1}{4}+\ln\left(\frac{4}{5}\right)\\ &\approx 0.026856\end{align}$$
More Reading:
Lerch Transendent
Lerch Identity
Polylogarithm
