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?
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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
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Thanks bhattacharjee! The special case you provided in the link is useful! – Arbiturka Apr 16 '18 at 14:11
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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
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Stars and bars 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
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Please give more context. 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. 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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