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 am trying to prove how $$g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}$$

or $\sum ak(k-1)r^(k-1) = 2a(1-r)^{-3}$.

I don't know what I am doing wrong and am at my wits end.

The attempt at a solution (The index of the summation is always $k=2$ to infinity)

\begin{align*} \sum ak(k-1)r^{k-1} &=a \sum k(k-1)r^{k-2}\\ &=a \sum (r^k)''\\ &=a \left(\frac{r^2}{1-r}\right)'' \end{align*}

From this point I get a mess, and the incorrect answer. The thing I have a problem is I think $\sum (r^k)''$ when $k$ is from $2$ to infinity is $r^2/(1-r)$, since the first term in this sequence is $r^2$. I don't think that is correct though.

share|cite|improve this question

$$f(x)=\sum_{k=0}^\infty ax^k=\frac{a}{1-x}\;\;,\;\;|x|<1$$

$$f'(x)=\sum_{k=1}^\infty akx^{k-1}=\frac{a}{(1-x)^2}\;\;,\;\;|x|<1$$

$$f''(x)=\sum_{k=2}^\infty ak(k-1)x^{k-2}=\frac{2a}{(1-x)^3}\;\;,\;\;|x|<1$$

Withing the convergence interval, you can derivate/integrate elementwise.

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.