Find the sum $\sum_{n=1}^\infty \frac{n^2}{3^n}$

Question

This problem comes from UChicago's Math Subject GRE Practice Problems. Specifically week 1, problem 6:

Evaluate the sum $\sum_{n=1}^\infty\frac{n^2}{3^n}$.

The answer says "Use the standard trick." Any idea what they mean? I know the sum of a geometric series, but I'm not sure what to do here given the $n^2$.

Answer 1

Here, we use concept of convergence of power series. for $|x|<1$,

$\sum{x^n}= \frac{x}{1-x}$ Since ,Rad of convergence of power series is 1 ,we can differentiate term by term in its interval of convergence

$\sum{nx^{n-1}}$ = $\frac{1}{(1-x)^2}$

$\sum{nx^n}=\frac{x}{(1-x)^2}$

Again differentiating, $\sum{n^2x^{n-1}}=\frac{1+x}{(1-x)^3}$

$\sum{n^2x^n}=\frac{x^2+x}{(1-x)^3}$

Since, $x=1/3$ lies in interval of convergence We get, $\sum\frac{n^2}{3^n}=\frac{\frac{1}{9}+\frac{1}{3}}{\frac{2^3}{3^3}}$ $=\frac{3}{2}$

Answer 2

You need to realize that this series is the value of the power series $\sum_{n \geq 1} n^2x^n$ evaluated $x = 1/3$, and this power series can be computed as a rational function of $x$ on its interval or disc of convergence by repeated differentiation starting from the geometric series $$ \sum_{n \geq 1} x^n = \frac{x}{1-x} $$ for $|x| < 1$. More precisely, apply $d/dx$ to both sides of that equation, then multiply by $x$, do those again, and then set $x = 1/3$.

You are expected to realize immediately that coefficient factors of $n$ get introduced in power series by differentiating and then multiplying by $x$: $$ x\frac{d}{dx}\sum_{n \geq 0} a_nx^n = \sum_{n \geq 0} na_nx^n = \sum_{n \geq 1} na_nx^n. $$ So for each positive integer $k$, $$ \left(x\frac{d}{dx}\right)^k\sum_{n \geq 0} a_nx^n = \sum_{n \geq 1} n^ka_nx^n. $$

Remark. Differentiating twice without multiplying by $x$ between the two derivatives will not lead directly to the desired answer, but it can be made to work by remembering prior calculations. When $|x| < 1$, $$ \sum_{n \geq 1} x^n = \frac{x}{1-x} \Longrightarrow \sum_{n \geq 1} nx^{n-1} = \frac{1}{(1-x)^2} $$ by taking derivatives of both sides. Then taking derivatives again, $$ \sum_{n \geq 1} n(n-1)x^{n-2} = \frac{2}{(1-x)^3}, $$ so $$ \sum_{n \geq 1} n^2x^{n-2} = \frac{2}{(1-x)^3} + \sum_{n \geq 1} nx^{n-2}. $$ Now multiply both sides by $x^2$ and recall the earlier formula for $\sum_{n \geq 1} nx^{n-1}$: \begin{align*} \sum_{n \geq 1} n^2x^{n} & = \frac{2x^2}{(1-x)^3} + x\sum_{n \geq 1} nx^{n-1} \\ & = \frac{2x^2}{(1-x)^3} + \frac{x}{(1-x)^2}. \end{align*} Now set $x = 1/3$: $$ \sum_{n \geq 1} \frac{n^2}{3^n} = \frac{2/9}{8/27} + \frac{1/3}{4/9} = \frac{3}{4} + \frac{3}{4} = \frac{3}{2}. $$

Answer 3

Another form of the trick is $$n^2 x^n=n(n-1) x^n+n x^n= x^2\big[n(n-1)x^{n-2}\big]+x \big[n x^{n-1}\big]$$ that is to say $$n^2 x^n=x^2 (x^n)''+x(x^n)'$$