Asymptotically equivalent series for uniform convergence

Question

I have to find sets of uniform convergence of $$\sum_{n=1}^{\infty}n^2 \sin \frac{x}{n^4}$$ what if I study this series passing to the asymptotically equivalent $$\sum_{n=1}^{\infty}n^2 \frac{x}{n^4}$$ ?

In this case I could use the power series theorem!

Is it correct?

Answer 1

$$ f_n(x):=n^2\sin{x\over n^4}\;\forall\; n $$

Claim: $\;0<M<\infty\implies\sum_{n=1}^{\infty}f_n$ converges uniformly in $[-M,M]$.

Proof: Suppose $x\in[-M,M]$. $$ |\sum_{n=p}^q f_n(x)|\le\sum_{n=p}^q|f_n(x)|\le\sum_{n=p}^q n^2{|x|\over n^4}\le\sum_{n=p}^q{M\over n^2}=M\sum_{n=p}^q {1\over n^2}$$ So given $\varepsilon>0$ we can find $k(\varepsilon):\forall\;p\ge q\ge k(\varepsilon)\implies$ $$ |\sum_{n=p}^q f_n(x)|<\varepsilon\;\forall\; x\in[-M,M] $$ as $\sum_{n=1}^\infty{1\over n^2}$ converges. So by Cauchy's criterion we're done.

Claim: $\sum_{n=1}^{\infty}f_n$ doesn't converge uniformly on $[0,\infty[$

Proof: Suppose our claim is not true. So $\exists k\in\mathbb{N}:p\ge q\ge k\implies$ $$ |\sum_{n=p}^q f_n(x)|<1\;\forall\; x\in[0,\infty[ $$ Fix $x={k^4\pi\over2}\in[0,\infty[$. Letting $p=q=k$ we see $$ |\sum_{n=p}^q f_n(x)|=k^2\ge1 $$ which is a contradiction.

Similarly, $]-\infty,0]$ is not a set of uniform convergence.