Good afternoon to everyone, I have the following question:
What does the arithmetico-geometric series:
$$S = \sum^{\infty}_{n=1} ne^{-nrt}$$
Converge to?
($r > 1$, $t > 1$)
I tried to break it up in the series of the even number and the series of the odd numbers multipliers: \begin{align*} S &= e^{-rt} + 2e^{-2rt} + ... \\ \\ S &= S_1 + S_2 \\ \\ S_1 &= (3-2)e^{-rt} + (3-0)e^{-3rt} + (3+2)e^{-5rt} + ... \\ \\ &= \frac{3e^{-rt}}{1-e^{-2rt}} -2e^{-rt} +\frac{2e^{-5rt}}{1-2e^{-2rt}} \\ \\ S_2 &= 2e^{-2rt} + 4e^{-4rt}+... \\ \\ S_2 &= \frac{2e^{-2rt}}{1-2e^{-2rt}} \end{align*}
Hence,
$$ S =\frac{2(e^{-5rt}+e^{-2rt})}{1-2e^{-2rt}} + \frac{3e^{-rt}}{1-e^{-rt}}-2e^{-rt} $$
But honestly, I'm not sure at all... Plus I bet $1\$$ that I made some computational mistake...
Ok, now I see it. I've never integrated / differentiated a series before...
\begin{equation*} S = \sum^{\infty}_{n=1} ne^{nrt} \\ \end{equation*}
Let $e^{-rt}=x$.
\begin{align*} S &= \sum^{\infty}_{n=1} nx^n\\ \\ &= x\sum^{\infty}_{n=1} nx^{n-1} \\ \\ &= x\sum^{\infty}_{n=1} \frac{\partial S_1}{\partial x} \end{align*}
It follows that it must be:
\begin{align*} S_1 &= \sum^{\infty}_{n=1} x^n &= \frac{x}{1-x} \end{align*}
Hence,
\begin{equation*} \frac{\partial S_1}{\partial x} = \frac{1}{(1-x)^2} \end{equation*}
Therefore,
\begin{align*} S &= \frac{x}{(1-x)^2} \\ \\ &= \frac{e^{-rt}}{(1-e^{-rt})^2} \end{align*}
You can see that $$ \sum_{n=1}^\infty nx^n = \sum_{n=1}^{\infty}\sum_{k=1}^n x^n = \sum_{k=1}^\infty\sum_{n=k}^\infty x^n = \sum_{k=1}^\infty \frac{x^k}{1-x} = \frac{x}{(1-x)^2}. $$ Now it is just a matter of replacing $x$ with $e^{-rt}$.