<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] 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 $r=\dfrac{1}{5}$.
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