Evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n} $ The following series converges to 3/2 but I do not know why.
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} = \frac{3}{2} $$
Searching via Google did not yield anything useful. I'm wondering if there's some sort of algorithm that can be used to solve this, or better yet, a closed-form equation for evaluating infinite sums that are in a form similar to
$$ \sum_{n=1}^\infty \frac{n^a}{b^n} $$
Or maybe it's just a Taylor series that I don't recognize.
 A: Hint:
$$\sum_{n=2}^\infty n(n-1)  x^{n-2}
= \frac {d^2}{dx^2} \sum_{n=0}^\infty  x^n
= \frac {d^2}{dx^2} \frac 1{1-x} = \frac 2{(1-x)^3} 
$$
$$\sum_{n=1}^\infty n  x^{n-1}
= \frac {d}{dx} \sum_{n=0}^\infty  x^n
= \frac {d}{dx} \frac 1{1-x} = \frac 1{(1-x)^2} $$
A: Jonquière's function, also known as the polylogarithm, is defined as a power series:
$$\mathrm{Li}_s(z)=\sum_{k=1}^{\infty}\dfrac{z^k}{k^s}$$
Your sum can therefore be evaluated as $\mathrm{Li}_{-2}(\frac{1}{3})$. Fortunately for us, explicit expressions of $\mathrm{Li}_{s}(z)$ are known, for integer values of $s$ at least. And in particular:
$$\mathrm{Li}_{-2}(z)=\dfrac{z(1+z)}{(1-z)^3}$$
such that:
$$\mathrm{Li}_{-2}\left(\frac{1}{3}\right)=\dfrac{\frac{1}{3}\times\frac{4}{3}}{\frac{8}{27}}=\dfrac{3}{2}$$
A: A solution using only the fact that $\sum_{n=0}^\infty q^n = \frac{1}{1-q}$ when $|q|<1$:
Let $S_1 = \sum_{n=1}^\infty\dfrac{n^2}{3^n}$, then $3S_1 = \sum_{n=0}^\infty\dfrac{(n+1)^2}{3^n}$, so we have $$2S_1 = 3S_1 - S_1 = 1 + \sum_{n=1}^\infty\dfrac{2n+1}{3^n} = 1 + 2S_2 + \sum_{n=1}^\infty\dfrac{1}{3^n} = 1 + 2S_2  + \frac{1}{2}$$
where $S_2 = \sum_{n=1}^\infty\dfrac{n}{3^n}$
Similarly we have $$2S_2 = 3S_2 - S_2 = \sum_{n=0}^\infty\dfrac{n+1}{3^n} - \sum_{n=1}^\infty\dfrac{n}{3^n} = \sum_{n=0}^\infty\dfrac{1}{3^n} = \frac{3}{2}$$
So $S_1 = \dfrac{1+\frac{3}{2}+ \frac{1}{2}}{2} =\dfrac{3}{2}$
A: Hint : Consider the series $\sum _{n\geq 1}3^{-n}x^n$ and differentiate it twice.
