To find closed form of given sum

Question

Can you guys give me hint or some start for this question. It looks easy but I am not getting. Question is to find closed form ( in terms of x and n only) of following sum. $\sum_{k=1}^n ke^{kx} $ ...where $n\in \mathbb{N}$ and $x\in \mathbb{R}-\{0\}$

Answer 1

Note that if $y=e^{kx}$ then $y'=ke^{kx}$ (derivative with respect to $x$)

That should be sufficient hint, I think.

Answer 2

Note that for $f(x)=e^{kx}$ we have $f'(x)=ke^{kx}$.

Then $\sum_{k=1}^n ke^{kx}=\sum_{k=1}^n \frac{d}{dx} e^{kx}=\frac{d}{dx}\sum_{k=1}^n e^{kx}\stackrel{\text{geometric sum}}{=}\frac{d}{dx} \dfrac{e^x(e^{nx}-1)}{e^x-1}$

Answer 3

Mark Bennet's hint is the best for your question. I present here a different method that I believe is worth learning.

If you try to do the sum for small values of $n$, you'll see that one way to do it is to regroup terms:

For $k=1$, you have a mere $e^x$.

For $k=2$, you have $2e^{2x}$. To have what resumbles a geometric sum, you'd write it as $e^{2x}+e^{2x}$ and then write the sum so far as $(e^x+e^{2x})+e^{2x}$.

For $k=3$, you have $3e^{3x}$, so here again write it as $e^{3x}+e^{3x}+e^{3x}$, and the sum becomes $(e^x+e^{2x}+e^{3x})+(e^{2x}+e^{3x})+e^{3x}$. You see that you're accumulating geometric sums, starting in a later term each time. You can better see this if you organize all the terms of your sum in the following way:

\begin{align*} &e^x & & & \\ &e^{2x} & e^{2x} & &\\ &e^{3x} & e^{3x} && e^{3x}\\ &e^{4x} & e^{4x} && e^{4x} && e^{4x}\\ \end{align*} and so on.. Your original sum is as if you're summing row by row. But when you sum column by column, you get geometric sums.

You can achieve this formally as follows:

\begin{align*} \sum_{k=1}^nke^{kx}&=\sum_{k=1}^n\sum_{l=k}^ne^{lx}\\ &=\sum_{k=1}^n\sum_{l=k}^n(e^x)^k. \end{align*}

I leave the rest to you.