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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.

Community
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leqs
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2 Answers2

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$$\begin{align} \sum\limits_{k=1}^{\infty}kx^k&=\sum_{k=1}^{\infty}(k+1)x^k-\sum_{k=1}^{\infty}x^k\\ &=\left(\sum^{\infty}_{k=1}x^{k+1}\right)^{\prime}-\frac{x}{1-x}\\ &=\left(\frac{x^2}{1-x}\right)^{\prime}-\frac{x}{1-x}\\ &=\frac{x}{(1-x)^2} \end{align}$$ Here $|x|&lt1$.

Daoyi Peng
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As you know $\sum_{k=0}^\infty x^k$ you can differentiate the result. Justify that you can differentiate the series term-by-term.

Jochen Wengenroth
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  • Oh, differentiation of course. – leqs Apr 24 '12 at 14:01
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    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