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.
 A: 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
A: We know the GP progression formula for when common ratio less than $1$ which is pretty simple to derive and is $\displaystyle \sum_{k=1}^\infty r^{k} = \dfrac{1}{1-r} $ from there we can write the following,
\begin{align} \int \sum\limits_{k=2}^\infty r^{k-1} dr = \int \frac{1}{1-r} dr \end{align}
After integration we get:
\begin{align} \sum\limits_{k=2}^\infty \frac{r^{k}}{k}=-\ln(1-r)\end{align}
Now start from $k=2$ instead of $k=1$ and let $r=\dfrac{1}{5}$.
A: 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 $x=\frac15$:
$$f\left(\frac15\right)=\ln \left(1-\frac15\right)+\frac1{1-\frac15}-1=\ln \frac45-\frac14.$$
A: Hint
For the time being, forget the $\frac 15$ and consider that what you need is
$$S=\sum_{i=1}^\infty \frac i {i+1} x^{i+1}$$ Derive with respect to $x$
$$S'=\sum_{i=1}^\infty i x^{i}=x \sum_{i=1}^\infty i x^{i-1}=x\left(\sum_{i=1}^\infty  x^{i}\right)'=x \times \left(\frac x {1-x}\right)'=\frac{x}{(1-x)^2}$$ Now, integration (by parts) to get $S$ and, when done, make $x=\frac 15$.
A: Thanks all who have given the answers. Some people use "Lerch Transenden" methods to solve some uses "integration", "convergence" etc.  But I was looking to solve it as using 10+2 level math techniques. 
Finally I got success in solving it. See my solution below -
= (1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4 +.........
= (1- 1/2)(1/5)^2 + (1- 1/3)(1/5)^3 + (1- 1/4)(1/5)^4 +.........
= (1/5)^2+(1/5)^3+(1/5)^4+.......−{1/2(1/5)^2+1/3(1/5)^3+1/4(1/5)^4+.....}
= (1/5)+(1/5)^2+(1/5)^3+(1/5)^4+.......−{(1/5)+1/2(1/5)^2+1/3(1/5)^3+1/4(1/5)^4+.....}
= (1/5)/(1- 1/5) - {-log(1- 1/5)}
= (1/5)/(4/5) + log(4/5)
= 1/4 + log(4/5)
Here log(1-x) = -x -x^2/2 -x^3/3 -..........
Once again thank you all of you.  
