2
$\begingroup$

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.

  • 0
    thanks, @Brian answered it2012-10-23

1 Answers 1

4

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 $x\ne 1$.

  • 0
    @Vaolter: You’re welcome.2012-10-23