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.

It is given:
$$\frac12\cdot \left(\frac15\right)^2+\frac23\cdot \left(\frac15\right)^3+\frac34\cdot \left(\frac15\right)^4+\cdots+\frac{n}{n+1}\cdot \left(\frac15\right)^{n+1}+\cdots$$
Consider the function:
$$f(x)=\frac12\cdot x^2+\frac23\cdot x^3+\frac34\cdot x^4+\cdots+\frac{n}{n+1}\cdot x^{n+1}+\cdots$$
We want to find $f\left(\frac15\right)$. (Why?)
Take derivative from both sides:
$$f'(x)=\color{blue}{x+2x^2+3x^3+\cdots +nx^n+\cdots}$$
Consider another function:
$$g(x)=1+x+x^2+x^3+\cdots+x^n+\cdots=\frac1{1-x}, \quad \left(x=\frac15<1\right)$$
Take derivative from both sides:
$$g'(x)=1+2x+3x^2+\cdots+nx^{n-1}+\cdots=\frac1{(1-x)^2}$$
Multiply both sides by $x$:
$$xg'(x)=\color{blue}{x+2x^2+3x^3+\cdots+nx^n+\cdots}=\frac{x}{(1-x)^2}$$
Note that $f'(x)=xg'(x)$, so:
$$f'(x)=\frac{x}{(1-x)^2}$$
Now we integrate both sides:
$$\int f'(x)dx=\int \frac{x}{(1-x)^2}dx \Rightarrow\\
f(x)=\int \frac{x-1+1}{(1-x)^2}dx=-\int \frac1{1-x}dx+\int \frac1{(1-x)^2}dx =\\
\ln (1-x)+\frac1{1-x}+C$$
Note that $f(0)=0$:
$$f(0)=\ln (1-0)+\frac1{1-0}+C=0 \Rightarrow C=-1$$
So:
$$f(x)=\ln (1-x)+\frac1{1-x}-1$$
At last we plug $x=\frac15$:
$$f\left(\frac15\right)=\ln \left(1-\frac15\right)+\frac1{1-\frac15}-1=\ln \frac45-\frac14.$$
