Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been struggling with this functional series. $$\sum_{n=1}^{\infty}{(-1)^{n-1}n^2x^n}$$ I need to calulate the sum.
Any tips would be appreciated.

share|cite|improve this question
This sort of sum can be done by "differentiating/integrating under the sum". Try differentiating or integrating each term with respect to $x$ to get rid of the factors of $n$; if it doesn't work out as is, you might need to pull some power of $x$ outside the sum first. Then when you've gotten rid of the $n^2$, you can evaluate the sum, and then you can do the inverse of the integration/differentiation that you did before to get the final result. – joriki Mar 21 '11 at 17:44
up vote 7 down vote accepted

Try considering what happens when you differentiate the following with respect to $x$:
$1/(1+x) = 1-x+x^2-x^3+x^4...$

That should get you thinking in the right direction.

share|cite|improve this answer

Note that $n^2=n(n-1)+n$ and hence the given series $\displaystyle\sum\limits_{n\geq 1} n^2 x^n=: f(n;x)$ can be written as

$f(n;x)=\displaystyle\sum_{n\geq 1} n(n-1)x^n+\displaystyle\sum_{n\geq 1} n x^n=f_1(n;x)+f_2(n;x)$ respectively. Let us determine these $f_1(n;x)$ and $f_2(n;x)$ separately.

Now $f_1(n;x)=x^2 \displaystyle\sum_{n\geq 1} n(n-1)x^{n-2}=x^2 \frac{2}{(1+x)^3}=\frac{2x^2}{(1+x)^3}$ by using the fact $(1+x)^{-1}=\displaystyle\sum_{n\geq 1} x^n$ and two times successive differentiation.

On the other hand $f_2(n;x)=-x \frac{1}{(1+x)^2}=\frac{-x}{(1+x)^2}$.

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

share|cite|improve this answer


If you divide your series by $x$ and then integrate it, it might be a little simpler.

Then ask yourself what you can do to make it even simpler. If you are lucky you will eventually end up with a geometric series where you know the answer.

Then undo all the steps you have taken (in reverse order), and you should have a solution to your original question.

share|cite|improve this answer

Incorporate $-1$ into $x$. Now consider the derivatives of $\sum z^n$. How can you get $n^2$ to show up?

share|cite|improve this answer

The ratio of convergence is $$\limsup_{n\to\infty}\frac{1}{\sqrt[n]{n^2}}=1,$$ so the series in convergent on $(-1;1)$. Now if move one $x$ factor before the sum, so that the exponent of $x$ in the sum is $n-1$, what happens if you integrate elementwise? Repeat!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.