How to find sum of the power series $\sum_{n=1}^{\infty} x^2 e^{-nx}$?

Question

I got this power series $$\sum_{n=1}^{\infty} x^2 e^{-nx} $$ And i need to prove uniform convergence when $x \in[0;1]$ and find this sum. I have proven uniform convergence, but i have no idea how to find this sum. It's clear, that $\sum_{n=1}^{\infty} x^2 e^{-nx}$ is a geometric series $\frac{x^2 e^{-x}}{1-e^{-x}}$. Is it correct that $\sum_{n=1}^{\infty} x^2 e^{-nx}$ equals $\int_0^1 \frac{x^2 e^{-x}}{1-e^{-x}}$?

Answer 1

You already computed the correct answer

$$\sum_{n=1}^\infty x^2e^{-nx}=x^2\frac{e^{-x}}{1-e^{-x}}.$$

You seem to conflate a few different concepts here: For one,

$$\sum_{n=1}^\infty x^2e^{-nx}=\lim_{k\rightarrow\infty}\sum_{n=1}^k x^2e^{-nx}$$

(a series is the limit of its partial sums) is the definition of a series.

Then, you have some "similarities" between series and integrals that do not factor into this problem: For starters, the Riemann integral approach:

$$\sum_{i=1}^N f(x_i)\Delta x_i\rightarrow\int_a^bf(x)\text dx$$

for a partition $a=x_1<...<x_N=b$ of an interval $[a,b]$, with $\Delta x_i=x_{i+1}-x_i$. The limit converges for any sequence of partitions as long as the mesh size $\max_{i=1,...,N-1}(x_{i+1}-x_i)$ converges towards zero.

Then, there is the integrability criterion / integral test that lets you determine whether a series converges at all by comparing it to the integral over its summation sequence. That only holds for non-negative monotonely decreasing functions, though, and is a comparison, not an equality.

Again, none of this has anything to do with the answer to your question, but might clear up some concepts you seem to have difficulty grasping.