# Find value of infinite sum [duplicate]

Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$

How would I go about deriving the value of the following infinite sum: $\sum\limits_{k=1}^\infty kx^k$ ?

I thought about expanding first: $\sum\limits_{k=1}^\infty kx^k= x + 2x^2 + 3x^3 + \cdots$

Then a bit of algebra: $\sum\limits_{k=1}^\infty kx^k - \sum\limits_{k=1}^\infty (k-1)x^k = x + x^2 + x^3 + \cdots + 1 -1$

And now I'm stuck with this: $\sum\limits_{k=1}^\infty x^k = \frac{x}{1-x}$

How can I introduce the $k$ into $\sum\limits_{k=1}^\infty x^k$ ? Or is there a different approach that I don't know of?

Any help is much appreciated.

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## marked as duplicate by Pedro Tamaroff♦, Did, Hans Lundmark, J. M., Zev ChonolesMay 13 '12 at 16:37

See here for ideas. – David Mitra Apr 24 '12 at 13:55
Thanks @David. I got the idea from Jonas Meyer's answer. Briefly, I would rewrite: $\sum\limits_{k=1}^\infty kx^k = x(1 + x + x^2 + ...) + x^2(1 + x^2 + ...) + ... = x(1+ x + x^2 + ...)^2$. Then $\sum\limits_{k=1}^\infty kx^k = \frac{x}{(1-x)^2}$. – leqs Apr 24 '12 at 14:27
There must be at least five duplicates of this question... – Hans Lundmark Apr 24 '12 at 14:29
@Hans Indeed. I voted to close. – Pedro Tamaroff Apr 24 '12 at 16:09

As you know $\sum_{k=0}^\infty x^k$ you can differentiate the result. Justify that you can differentiate the series term-by-term.
Thanks @Jochen Wengenroth. I differentiate both sides here: $\sum\limits_{k=1}^\infty x^k = \frac{x}{1-x}$. Then I get: $\sum\limits_{k=1}^\infty kx^(k-1) = \frac{1}{(1-x)^2}$. I then multiply both sides by $x$ and I get my answer. – leqs Apr 24 '12 at 14:21
\begin{align} \sum\limits_{k=1}^{\infty}kx^k&=\sum_{k=1}^{\infty}(k+1)x^k-\sum_{k=1}^{\infty}x^k\\ &=\left(\sum^{\infty}_{k=1}x^{k+1}\right)^{\prime}-\frac{x}{1-x}\\ &=\left(\frac{x^2}{1-x}\right)^{\prime}-\frac{x}{1-x}\\ &=\frac{x}{(1-x)^2} \end{align} Here $|x|<1$.