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How can I sum this infinite series $\sum_{n=1}^\infty n2^{n-1}$?. It looks like its a derivative of some function of 2 but have not made progress with that approach or may there is a result that I don't know which I must make use of.

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Wouldn't this be $\infty$? –  gt6989b Oct 23 '12 at 16:18
    
not sure, how about summing upto 100, can we get an expression for it in this case? –  Vaolter Oct 23 '12 at 16:25
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@Vaolter If the sum is finite, then it does converge. Do you know the Geometric Series? –  Pragabhava Oct 23 '12 at 16:33
    
In the usual metric, it diverges. In the $2$-adic metric it converges, and Brian's formula gives you the value. See en.wikipedia.org/wiki/P-adic_number –  GEdgar Oct 23 '12 at 16:56
    
thanks, @Brian answered it –  Vaolter Oct 23 '12 at 17:15
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1 Answer 1

up vote 4 down vote accepted

You’re right that this can be seen as involving a derivative, but you’ve overlooked the obvious: $\sum_{n\ge 1}n2^{n-1}$ diverges (and does so rather rapidly, at that). It’s the $2$ that louses things up. Let’s look at the same series with an unspecified number in place of the $2$.

Let $f(x)=\sum_{n\ge 0}x^n$; this is just a geometric series, so $f(x)=\frac1{1-x}$. Now differentiate:

$$f\,'(x)=\sum_{n\ge 1}nx^{n-1}=\frac1{(1-x)^2}\;.$$

However, this is meaningful only if $x$ is in the interval of convergence (or if you’re dealing with formal power series). Here the interval of convergence is $(-1,1)$, which does not contain $2$.

If you want only a finite sum, you can use the same basic idea. Let $$f(x)=\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}\;;$$ then

$$\begin{align*} \sum_{k=1}^nkx^{k-1}&=f\,'(x)=\frac{-(n+1)x^n(1-x)+1-x^{n+1}}{(1-x)^2}\\ &=\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2}\;, \end{align*}$$

a formula which is clearly valid for all $x\ne 1$.

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thanks @Brian you have answered the question –  Vaolter Oct 23 '12 at 17:16
    
@Vaolter: You’re welcome. –  Brian M. Scott Oct 23 '12 at 17:21
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