Sequence sum question: $\sum_{n=0}^{\infty}nk^n$

Question

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

Can anybody help me?

Answer 1

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?

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $$\begin{align*} \lim_{m\to\infty}S_m &= k+k\lim_{m\to\infty} S_{m-1}+\frac{k^2}{1-k}-\lim_{m\to\infty}\frac{k^{m+1}}{1-k}\\ l &= k+kl+\frac{k^2}{1-k}-0\\ (1-k)l &= k+\frac{k^2}{1-k}\\ l&= \frac{k}{1-k}+\left( \frac{k}{1-k} \right)^2\\ l&=\frac{k}{(1-k)^2}. \end{align*}$$

Answer 6

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

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