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I want to know the limit of the series $$ \sum_{k=1}^{\infty} k^2(1-q)^k,$$ where $0<q<1.$ I have checked from the criterium $a_{k+1}/a_{k} < 1$ that the series converges but I don't know hot to compute a limit or an upper bound for this limit?

Arbiturka
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  • https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 – lab bhattacharjee Apr 16 '18 at 14:09
  • Thanks bhattacharjee! The special case you provided in the link is useful! – Arbiturka Apr 16 '18 at 14:11
  • You may simply apply $\left(x\cdot\frac{d}{dx}\right)^2$ to $\frac{x}{1-x}$ and evaluate such expression at $x=1-q$. – Jack D'Aurizio Apr 16 '18 at 14:18
  • [Stars and bars](https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)) provide an alternative derivation, based on $\sum_{k\geq 2}\binom{k}{2}x^k = \frac{x^2}{(1-x)^3}$, $\sum_{k\geq 1}\binom{k}{1}x^k = \frac{x}{(1-x)^2}$ and $k^2=2\binom{k}{2}+\binom{k}{1}$. – Jack D'Aurizio Apr 16 '18 at 14:20
  • Please give more [context](http://meta.math.stackexchange.com/a/9960). Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please [avoid "I have no clue" questions](https://math.meta.stackexchange.com/a/27933). Defining keywords and trying a simpler, similar problem often helps. – robjohn Jul 30 '18 at 22:00

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Let $x\in (-1,1)$ and: $$ f(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x} $$ Furthermore: $$ f'(x)=\sum_{k=1}^\infty kx^{k-1}=\sum_{k=0}^\infty (k+1)x^k = \frac{1}{(1-x)^2} $$ and $$ f''(x)=\sum_{k=1}^\infty (k^2+k)x^{k-1} = -\frac{2x-2}{(1-x)^4} $$ Then, your sum is: $$ (1-q)(f''(1-q)-f'(1-q)) $$

Bill O'Haran
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