Does the series of functions $\sum_{n=0}^\infty f_n$ with $f_n: [1,2] \to \mathbb{R}: x \mapsto xe^{-nx}$ converge pointwise? Does it converge uniformly?

My attempt:

First notice that $$\left\Vert \frac{1}{e^{nx}}\right\Vert_{\infty} = \sup_{x \in [1,2]}|e^{-nx|} \leq e^{-n}$$

By comparing to the convergent geometric series $\sum e^{-n}$, we deduce by Weierstrass' $M$-test that $\sum_{n=0}^\infty e^{-nx}$ converges uniformly.

Now, because the sequence of functions $(x)_n$ is constant, it is decreasing (not strictly, but this is not necessary), and it is also uniformly bounded (because $\sup_{x \in[1,2]}|x| = 2)$. Hence, by Abel's criterion for uniform convergence we deduce that the series

$$\sum_{n=0}^\infty x e^{-nx} = \sum_{n=0}^\infty f_n$$ converges uniformly, as desired.

Is this correct? Are there easier approaches?


1 Answer 1


It is correct, but there is no need to use Abel's criterion for uniform convergence. Just note that$$(\forall x\in[1,2])(\forall n\in\mathbb{N}):\bigl|xe^{-nx}\bigr|\leqslant2e^{-n}.$$Therefore, you can apply Weierstrass' $M$-test directly.


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