I know that $$\sum_{n=1}^\infty(-1)^n\frac{n^2}{3^n}$$ converges by alternating series test, but I can get sum of it.
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Check if my formatting preserves your question. Look up how to use Mathjax. Also, we don't know at what level you try to solve the problem. Do you know derivatives? – Andrei Apr 02 '20 at 01:24
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2https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 – lab bhattacharjee Apr 02 '20 at 01:58
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A couple of similar questions: How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$?, Proof of the equality $\sum\limits_{k=1}^{\infty} \frac{k^2}{2^k} = 6$, Finding the sum of series $\sum \limits_{n=1}^{\infty} (-1)^{n}\frac{n^2}{2^n}$. You could find some of them using Approach0 or ... – Martin Sleziak Apr 02 '20 at 05:53
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... SearchOnMath. Or you could search for more general $\sum_{n=1}^\infty n^2 x^n$: Approach0 and SearchOnMath. Checking frequent tab in the tag ([tag:sequences-and-series]) is also worth trying. Advice on searching: How to search on this site? – Martin Sleziak Apr 02 '20 at 05:54
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@dusday Oh! you forgot about your problem, it seems. – Z Ahmed Apr 02 '20 at 14:26
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$\textbf{Hint:}$ You can start with the series \begin{equation*} \sum_{k = 0}^{\infty}x^{k} = \frac{1}{1-x} \end{equation*} and proceed by differentiating both sides twice.
Zac
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It would be very surprising if you need to know the value of the sum of this series. At the point that you are learning various convergence and divergence tests, there are exactly three kinds of series you can evaluate: constant terms, geometric series, and telescoping series. You do not find a fourth way until you can compute Taylor series of functions and evaluate the functions as a way to evaluate the series.
As an example of this fourth method... The Taylor series, centered at $0$, of $$ \frac{x(x-1)}{(x+1)^3} = \sum_{n = 1}^\infty (-1)^n n^2 x^n \text{.} $$ It has radius of convergence $1$. Evaluate at $x = 1/3$ to get your sum.
Eric Towers
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If $x|<1$, then $$\sum_{k=0}^{\infty}(-1)^k x^k =\frac{1}{1+x}$$ D. w.r.t. $x$, to get an multiply both sides by $x$ $$\sum_{k=0}^{\infty}(-1)^k ~k~ x^{k} =-\frac{x}{(1+x)^2}$$ Again differentiate w.r.t. $x$ and multiply by $x$ on both sides to get $$\sum_{k=0}^{\infty}(-1)^k ~k^2~ x^{k} =\frac{x(x-1)}{(1+x)^3},~ if~ |x|<1 $$ Finally put $x=1/3$, to get $$\sum_{n=1}^{\infty}(-1)^n \frac{n^2}{3^n}=\frac{-3}{32}$$
Z Ahmed
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Rewrite $$\sum_{n=1}^\infty(-1)^n\frac{n^2}{3^n}$$ as $$\sum_{n=1}^\infty n^2 x^n=\sum_{n=1}^\infty(n(n-1)+n)x^n=\sum_{n=1}^\infty n(n-1)x^n+\sum_{n=1}^\infty n x^n$$ that is to say $$x^2\sum_{n=1}^\infty n(n-1)x^{n-2}+x\sum_{n=1}^\infty n x^{n-1}=x^2\left(\sum_{n=1}^\infty x^n \right)''+x\left(\sum_{n=1}^\infty x^n \right)'$$
When done, make $x=-\frac 13$.
Claude Leibovici
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