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$$\sum _{r=1} ^{\infty} \frac{r^2}{(-5)^{r-1}}$$

The series comes out to be $$1^2 -\frac{2^2}{5} + \frac{3^2}{5^2}- \frac{4^2}{5^3} ........ $$

In series like these I try to break $T_r$ into two parts and only the first and the last term remain or try to combine $T_{r} + T_{r+1}$ and then break it. Wasn't able to reach anywhere doing that.

I even split the series into negative and positive but still wasn't able to resolve it.

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    Do you know how to evaluate $\sum\limits {r=1}^{\infty}r^2 z^r$ in general (for $|z| < 1$)? If not, you can derive the formula for it as follows: start with $\frac{1}{1-z} = \sum\limits{r=0}^{\infty}z^r$, then apply $z\frac{d}{dz}$ twice (i.e. differentiate with respect to $z$, then multiply by $z$, then repeat once). – Minus One-Twelfth Apr 03 '19 at 07:30
  • See https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 – lab bhattacharjee Apr 03 '19 at 08:47

2 Answers2

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$$\frac{1}{1-x}=\sum_{r=0}^{\infty}x^r$$ $$(\frac{1}{1-x})'=\sum_{r=1}^{\infty}rx^{r-1}$$ $$x(\frac{1}{1-x})'=\sum_{r=1}^{\infty}rx^{r}$$ $$(x(\frac{1}{1-x})')'=\sum_{r=1}^{\infty}r^2x^{r-1}$$ then let $x=-\frac{1}{5}$

E.H.E
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Hint $:$ Observe that $$\begin{align} \sum\limits_{n=1}^{\infty} n^2 x^n & = \sum\limits_{n=1}^{\infty} n(n-1)x^n + \sum\limits_{n=1}^{\infty} nx^n \\ & = \frac {2x^2} {(1-x)^3} + \frac {x} {(1-x)^2} \\ & = \frac {x(x+1)} {(1-x)^3} \end{align}$$

whenever $|x|<1.$

little o
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