Why does Weierstrass M-test not apply on $f_k:]-1,1[ \rightarrow \mathbb{R}, k \in \mathbb{N}, f_k(x)=x^2 e^{-kx^2}$?
It seems like it should, since one can display
$\forall k \ge 1, \forall x \in ]-1, 1[$
$$|f_k(x)| = |x^2e^{-kx^2}|=|x^2||e^{-kx^2}|\le|e^{-kx^2}|\le |e^{-k}| = \bigg (\frac{1}{e} \bigg )^k := M_k$$
Now
- $M_k > 0, \forall k \in \mathbb{N}$.
- $|f_k(x)| \le M_k$ is satisfied $\forall k \ge 1, \forall x \in ]-1, 1[$.
- $\sum_{k=1}^{\infty} M_k < \infty$ since $\sum_{k=1}^{\infty}\bigg( \frac{1}{e} \bigg)^k = \frac{1}{1-\frac{1}{e}}$ is a geometric series.
So it seems like all Weierstrass M-test's assumptions are met.
But in fact, $\sum_{k=1}^{\infty} f_k$ shouldn't converge uniformly, because e.g. for $\frac{1}{\sqrt{k}} \in ]-1,1[$
$$f_k(\frac{1}{\sqrt{k}})=(\frac{1}{\sqrt{k}})^2e^{-k(\frac{1}{\sqrt{k}})^2}=e(\frac{1}{k})$$
would, when summed, create a harmonic series, which diverges.