evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$

Question

I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$

I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.

Can anyone help me?

Thanks.

Answer 1

Here is a non-calculus method which works in these cases:

Put $$S = \frac{1}{3} + \frac{4}{9} + \frac{9}{27} + \frac{16}{81} + \dots$$

then $$\frac{1}{3}S = \frac{1}{9} + \frac{4}{27} + \frac{9}{81} + \dots$$

subtract:

$$\frac{2}{3}S = \frac{1}{3} + \frac{3}{9} + \frac{5}{27} + \dots$$

This has simplified the problem (since the numerators are now linear instead of quadratic), and if you repeat the process, you get the answer, using the sum of a geometric progression.

Answer 2

Hints:

For $\;|x|<1\;$ :

$$f(x):=\frac1{1-x}=\sum_{n=0}^\infty x^n\implies f'(x)=\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies$$

$$f''(x)=\frac2{(1-x)^3}=\sum_{n=2}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty n^2x^{n-2}-\sum_{n=2}^\infty nx^{n-2}\;\;\ldots$$

Answer 3

Here's a bit of a slick trick. Let's put $$S=\sum_{n=1}^\infty\frac{n^2}{3^n}$$ and $$L=\sum_{n=1}^\infty\frac{n}{3^n}.$$ You should be able to determine (via ratio test or some such) that both series are convergent. (You'll see why I brought up the other series in a moment.)

Note that $$\begin{align}S &= \frac13+\sum_{n=2}^\infty\frac{n^2}{3^n}\\ &= \frac13+\sum_{n=1}^\infty\frac{(n+1)^2}{3^{n+1}}\\ &= \frac13+\sum_{n=1}^\infty\frac{n^2}{3^{n+1}}+2\sum_{n=1}^\infty\frac{n}{3^{n+1}}+\sum_{n=1}^\infty\frac1{3^{n+1}}\\ &= \frac13+\frac13 S+\frac13 L+\frac13\sum_{n=1}^\infty\frac1{3^n}\\ &= \frac13 S+\frac13 L+\frac13\sum_{n=0}^\infty\frac1{3^n},\end{align}$$ so $$\frac23 S=\frac13 L+\frac13\sum_{n=0}^\infty\frac1{3^n},$$ so $$S=\frac12 L+\frac12\sum_{n=0}^\infty\frac1{3^n}.\tag{$\star$}$$ Now a similar trick shows us that $$\begin{align}L &= \frac13+\sum_{n=2}^\infty\frac{n}{3^n}\\ &= \frac13+\sum_{n=1}^\infty\frac{n+1}{3^{n+1}}\\ &= \frac13L+\frac13\sum_{n=0}^\infty\frac1{3^n},\end{align}$$ so $$\frac23 L=\frac13\sum_{n=0}^\infty\frac1{3^n},$$ so $$L=\frac12\sum_{n=0}^\infty\frac1{3^n},$$ and so by $(\star),$ we have $$S=\frac34\sum_{n=0}^\infty\frac1{3^n}.$$ All that's left is to evaluate the geometric series.

Answer 4

We have

\begin{gather*} S=\sum_{n=1}^{\infty}\frac{n^2}{3^n}=\frac{1}{3}\sum_{n=0}^{\infty}\frac{(n+1)^2}{3^{n-1}}=\frac{1}{3}\left(\sum_{n=0}^{\infty}\frac{n^2}{3^{n-1}}+2\sum_{n=0}^{\infty}\frac{n}{3^{n-1}}+\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}\right)=\frac{1}{3}\left(S+2 S_1+S_2\right). \end{gather*} Then $$ S_2=\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}=\frac{3}{2}, $$

$$S_1=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty \frac{n}{3^{n-1}}= \frac{1}{3}\sum_{n=0}^\infty \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty \frac{n}{3^{n}}+\sum_{n=0}^\infty \frac{1}{3^{n}}\right] =\frac{1}{3}\left[S_1+ S_2\right].$$

By solving this equation you get $S_1=\dfrac{3}{4}.$

Thus we have the equation $$ S=\frac{1}{3}(S+2 \cdot \frac{3}{4}+\frac{3}{2}). $$

By solving it you will get $S=\dfrac{3}{2}.$