Question

It's the series (from 1 to infinity) of k((2/3)^(k-1)). I know that the answer is 9, found from (1-2/3)^-2 but I have absolutely no idea where this comes from? Why is the sum of this series 9? Please help!

Answer 1

First, do you know
\[\sum_{k=0}^{N}r^N= \frac{ 1-r^{(N-1)} }{ 1-r }\]
see http://en.wikipedia.org/wiki/Geometric_progression#Derivation
for N= infinity, and 0<r <1, this is
\[ \sum_{k=0}^{\infty}r^k= \frac{ 1}{ 1-r } \]

If you know calculus, we can do the following:
\[ \sum_{k=0}^{\infty}r^k= 1+r+r^2+r^3+... \]
take the derivative with respect to r of both sides
\[ \frac{d}{dr}\left( \sum_{k=0}^{\infty}r^k \right)= 1+2r+3r^2+4r^3+...\]
notice that the series on the right hand side is your series. Swapping sides, we have
\[ \sum_{k=1}^{\infty}kr^{(k-1)} = \frac{d}{dr}\left( \sum_{k=0}^{\infty}r^k \right)\]

From the top part of this post we can replace the summation with 1/(1-r) to get
\[ \sum_{k=1}^{\infty}k\ r^{(k-1)} = \frac{d}{dr}\left(\frac{1}{1-r}\right)\]

now take the derivative of \( (1-r)^{-1} \). You get
\[ \frac{d}{dr}\left(\frac{1}{1-r}\right)= (1-r)^{-2} \text{ or } \frac{1}{(1-r)^2}\]
and the finally,

\[ \sum_{k=1}^{\infty}k\ r^{(k-1)} =\frac{1}{(1-r)^2}\]