Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

This question already has an answer here:

I am very confused about how to compute $$\sum_{n=0}^{\infty}nk^n.$$

Can anybody help me?

share|cite|improve this question

marked as duplicate by Martin Sleziak, Meta, Daniel Robert-Nicoud, Ali Caglayan, Zachary Selk Jun 17 '15 at 2:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I tried $\LaTeX$-ing your question, but I'm not sure this is what you meant. What is $k$? – Arturo Magidin Sep 25 '11 at 6:53
Yeah, you defintly have to provide some information about k, since that sum may actually diverge, for instance, if k was 1. – Chris Dugale Sep 25 '11 at 7:06
You already have a few answers using the derivative of geometric series; another possible approach would be summation by parts. – Martin Sleziak Sep 25 '11 at 7:26
Note that you can check the answers given bellow at wolframalpha: – Martin Sleziak Sep 25 '11 at 7:45
Yes, that's right. Thank you very much for the LATEXing. At this moment k is a constant with absolute value less than 1. But I am also curious about what results if k is larger than 1. – user16653 Sep 25 '11 at 17:29

A bit more generality gives $$ \begin{align} \sum_{n=k}^\infty\binom{n}{k}x^n &=\sum_{n=k}^\infty\binom{n}{n-k}x^n\\ &=\sum_{n=k}^\infty(-1)^{n-k}\binom{-k-1}{n-k}x^n\\ &=\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}x^{n+k}\\ &=x^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}x^n\\ &=x^k(1-x)^{-k-1} \end{align} $$ Using this with $k=1$ yields your formula with names changed. Therefore, $$ \sum_{n=1}^\infty nk^n=\frac{k}{(1-k)^2} $$ for $|k|<1$. If $|k|\ge1$, the series diverges.

share|cite|improve this answer

I assume k is some constant (presumably less than 1 in absolute value).

Then we calculate this as a derivative of the series $f(x) = \dfrac{1}{1-x} = \displaystyle \sum x^n$. Then $\displaystyle f'(x) = \dfrac{1}{(1-x)^2} = \sum nx^{n-1}$. So we can note that $xf'(x) = \displaystyle \sum nx^n$. Depending on how you want your indices, you may or may not need to add or take away a few terms. Also, this clearly only works for when the geometric series converges.

That's the idea. Is that what you wanted?

share|cite|improve this answer
I think the sign in the denominator of the derivative is -, not + – sigma.z.1980 Sep 25 '11 at 8:56
@sigma: Yep, it sure is. Thank you for that. – mixedmath Sep 25 '11 at 16:05

If you know the value of the geometric series $\sum\limits_{n=0}^{+\infty}x^n$ at every $x$ such that $|x|<1$ and if you know that for every nonnegative integer $n$, the derivative of the polynomial function $x\mapsto x^n$ is $x\mapsto nx^{n-1}$, you might get an idea (and a proof) of the value of the series $\sum\limits_{n=0}^{+\infty}nx^{n-1}$, which is $x^{-1}$ times what you are looking for.

share|cite|improve this answer

Asssuming $|k|<1$, this series converges uniformly and therefore sum and derivative can be interchanged:

$\sum_{n=0}^{\infty}nk^n=k \sum_{n=0}nk^{n-1} = k \sum_{n=0}^{\infty}\frac{d}{dk}(k^n) =k \frac{d}{dk} \sum_{n=0}^{\infty}k^n = k \frac{d}{dk}\frac{1}{1-k}=\frac{k}{(1-k)^2}$

EDIT: in case $k>1$ this is a completely different situation, as the geometric series diverges, but I don't know how to solve this problem.

share|cite|improve this answer
If $k>1$, then the given series converges since the terms don't approach $0$. – Michael Hardy Sep 25 '11 at 17:49
The fact that $\sum_{n=0}^\infty \frac{d}{dk} k^n = \frac{d}{dk} \sum_{n=0}^\infty k^n$ may also be remarked upon. For finite sums, this is routine. It doesn't hold for all infinite sums, but it does for power series in the interior of their intervals of convergence, and that is applicable here. – Michael Hardy Sep 25 '11 at 17:51
@Michael:isn't this interchange due to uniform convergence? – sigma.z.1980 Sep 25 '11 at 18:56
correction: in my first comment above, I menat of course "diverges". – Michael Hardy Sep 29 '11 at 14:11

By the ratio test your series converges for $|k|\lt 1$. Let $|k|\lt 1$, consider $$S_m=\sum_{n=0}^m nk^n.$$ Then $$\begin{align*} S_m &= 0 + k + \sum_{n=2}^m nk^n\\ &= k +\sum_{n=1}^{m-1}(n+1)k^{n+1}\\ &= k + k\sum_{n=1}^{m-1} nk^n + \sum_{n=1}^{m-1} k^{n+1}\\ &= k + k\sum_{n=1}^{m-1} nk^n + k^2\sum_{n=1}^{m-1} k^{n-1}\\ &= k + kS_{m-1} + k^2\sum_{n=0}^{m-2} k^n\\ &= k + kS_{m-1} + k^2\cdot \frac{1-k^{m-1}}{1-k}. \end{align*}$$ We know that $S_m\to l$ for some $l\in\mathbb{R}$. Taking limits in the last equality we get $$\begin{align*} \lim_{m\to\infty}S_m &= k+k\lim_{m\to\infty} S_{m-1}+\frac{k^2}{1-k}-\lim_{m\to\infty}\frac{k^{m+1}}{1-k}\\ l &= k+kl+\frac{k^2}{1-k}-0\\ (1-k)l &= k+\frac{k^2}{1-k}\\ l&= \frac{k}{1-k}+\left( \frac{k}{1-k} \right)^2\\ l&=\frac{k}{(1-k)^2}. \end{align*}$$

share|cite|improve this answer

$$\sum_{k=1}^n kp^k=\frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}$$

For a proof by induction and other methods, see here.

If $n\to \infty$, the series converges due to ratio test ($\lim_{n\to \infty}\left|\frac{(k+1)}{k}p\right|<1$), when $|p|<1$. You'll get $$ \sum\limits_{k=1}^\infty kp^k = \frac p{(p-1)^2}+\underbrace{\lim_{n\to \infty} \frac{np^{n+2}-(n+1)p^{n+1}}{(p-1)^2}}_{=0} = \frac p {(p-1)^2} $$

share|cite|improve this answer

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