Sum of a simple infinite series

Question

Evaluate:
$$\sum\limits_{n=1}^\infty \frac{n^2}{3^n}.$$

By the ratio test, $\displaystyle\lim_{n\to\infty} \frac{(n+1)^2}{3^{n+1}}\cdot\frac{3^n}{n^2}=1/3,$ which is less than 1, therefore the series is convergent.

Now I am stuck on how to evaluate this series, without the $n^2$ on top, it can be easily calculated by the geometric series formula. Any help would be appreciated.

See here for a solution technique; you'll have to differentiate twice, since you have $n^2$. — , Mar 15 '14 at 19:25
Write down the series for $1/(1-x)$. Differentiate. Multiply by $x$. Differentiate again. — David Mitra, Mar 15 '14 at 19:28
http://math.stackexchange.com/questions/139655/finding-closed-forms-for-sum-n-zn-and-sum-n2-zn — lab bhattacharjee, Mar 16 '14 at 03:25

score 4 · Accepted Answer · answered Mar 15 '14 at 19:46

Just to expand on David's comment: \begin{align*} \frac{1}{1-x} &= 1 + x + x^2 + \cdots + x^n + \cdots \\ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} &= 1 + 2x + 3x^2 + \cdots + nx^{n-1} + \cdots \\ \frac{x}{(1-x)^2} &= x + 2x^2 + 3x^3 + \cdots + nx^{n} + \cdots \\ \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \frac{1 + x}{(1 - x)^3} &= 1 + 2^2x + 3^2x^2 + \cdots + n^2x^{n-1} + \cdots \\ \frac{x(1 + x)}{(1 - x)^3} &= x + 2^2x^2 + 3^2x^3 + \cdots + n^2x^{n} + \cdots \\ \end{align*} Now plug in $x = \frac13$.

Thank you so much! I didn't know there is such a trick! – codeedoc Mar 15 '14 at 19:56 — codeedoc, Mar 15 '14 at 19:56

Sum of a simple infinite series

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