As part of a probability exam, we were required to compute the following sum:
$ S=\sum_{n=1}^{\infty}(\frac{2}{3})^{n-1}\frac{1}{3}e^{-\frac{2n(1-t)}{t}}$
Our lecturer likes finding probabilistic patterns which makes the computation possible, though here I cant see why he did what he did.
He claims that by setting: $p=(1-\frac{2}{3}e^{\frac{-2(1-t)}{t}})$, we can see that:
$S=\frac{1}{3}e^{-\frac{-2(1-t)}{t}}\frac{1}{p}\sum_{n=1}^{\infty}(1-p)^{n-1}p$
And then the right sum is simply 1 because it is a summation over the whole probability space of a geometric variable with parameter $p$.
Although I have the solution "right in front" of me, I am trying to learn from it but simply can't. The initial sum is simply too complicated and I can't think of a way I could see this pattern actually. Does anyone have any insights about that? maybe more points of view will help me understand how I should have got there myself.
Thanks.
If $r=\frac 2 3 e^{-\frac {2(1-t)} t}$ then the sum is $\frac 1 2 \sum\limits_{n=1}^{\infty} r^{n}$ which is a simple geometric series.