Sorry for the ill-framed question; I don't know how to better frame this. I'm trying to evaluate this summation:
$\sum_{n=1}^\infty n(5/6)^{n-1}$
So, the terms are : $1$, $2.(5/6)$, $3.(5/6)^2$, $4.(5/6)^3$, $5.(5/6)^4$ and so on...
Each term in this series can be seen as having an AP component in 1,2,3... and a GP component in (5/6), $(5/6)^2$, $(5/6)^3$...
I tried to break this up as a series of multiple GPs. That is, the first GP beginning with $(5/6)$ till $(5/6)^\infty$ , the 2nd beginning with $(5/6)^2$, the 3rd beginning with $(5/6)^3$ and so on.. The final summation would then be the sum of sums of each of the above GPs.
Is there a well-known method to solve summations of this kind? If yes, please point me to the appropriate sources. Else, please point me in the right direction of thinking..
Notice that for every complex number $z$ with $|z|<1$ we have $$ \sum_{n=1}^\infty nz^{n-1}=\frac{d}{dz}\sum_{n=0}^\infty z^n=\frac{d}{dz}\frac{1}{1-z}=\frac{1}{(1-z)^2}. $$ In particular for $z=5/6$ we have $$ \sum_{n=1}^\infty n\left(\frac56\right)^{n-1}=\left(1-\frac56\right)^{-2}=36. $$