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

Can anybody help me?

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  • $\begingroup$ I tried $\LaTeX$-ing your question, but I'm not sure this is what you meant. What is $k$? $\endgroup$ Sep 25, 2011 at 6:53
  • $\begingroup$ Yeah, you defintly have to provide some information about k, since that sum may actually diverge, for instance, if k was 1. $\endgroup$ Sep 25, 2011 at 7:06
  • $\begingroup$ You already have a few answers using the derivative of geometric series; another possible approach would be summation by parts. $\endgroup$ Sep 25, 2011 at 7:26
  • $\begingroup$ Note that you can check the answers given bellow at wolframalpha: wolframalpha.com/input/?i=sum%28n%2Ax%5En%29 $\endgroup$ Sep 25, 2011 at 7:45
  • $\begingroup$ 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. $\endgroup$
    – user16653
    Sep 25, 2011 at 17:29

6 Answers 6

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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.

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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?

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    $\begingroup$ I think the sign in the denominator of the derivative is -, not + $\endgroup$ Sep 25, 2011 at 8:56
  • $\begingroup$ @sigma: Yep, it sure is. Thank you for that. $\endgroup$
    – davidlowryduda
    Sep 25, 2011 at 16:05
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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.

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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.

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  • $\begingroup$ If $k>1$, then the given series converges since the terms don't approach $0$. $\endgroup$ Sep 25, 2011 at 17:49
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    $\begingroup$ 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. $\endgroup$ Sep 25, 2011 at 17:51
  • $\begingroup$ @Michael:isn't this interchange due to uniform convergence? $\endgroup$ Sep 25, 2011 at 18:56
  • $\begingroup$ correction: in my first comment above, I menat of course "diverges". $\endgroup$ Sep 29, 2011 at 14:11
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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 $$\begin{align*} \lim_{m\to\infty}S_m &= k+k\lim_{m\to\infty} S_{m-1}+\frac{k^2}{1-k}-\lim_{m\to\infty}\frac{k^{m+1}}{1-k}\\ l &= k+kl+\frac{k^2}{1-k}-0\\ (1-k)l &= k+\frac{k^2}{1-k}\\ l&= \frac{k}{1-k}+\left( \frac{k}{1-k} \right)^2\\ l&=\frac{k}{(1-k)^2}. \end{align*}$$

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

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 $$ \sum\limits_{k=1}^\infty kp^k = \frac p{(p-1)^2}+\underbrace{\lim_{n\to \infty} \frac{np^{n+2}-(n+1)p^{n+1}}{(p-1)^2}}_{=0} = \frac p {(p-1)^2} $$

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