<s>[INST] 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.
 [/INST] 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.  
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