Does the series $S(x) = \sum_{n=1}^\infty \frac{n^2x^2}{n^4+x^4}$ converge uniformly?

Question

does the above series uniformly converges? If it does, how to find the interval of x in which it uniformly converges? $ {(n^2-x^2)}^2 \ge 0 \Rightarrow n^4 + x^4 - 2n^2x^2 \ge 0$

This gave me $\frac{n^2x^2}{n^4+x^4} \le 1/2$. So, I couldn't apply Weierstrass M-Test as $\sum_{n=1}^\infty(1/2)$ diverges. Couldn't proceed further. Please help.

Answer 1

Note that for $x\in [-a,a]$ for any $a>0$, we have

$$\frac{n^2x^2}{n^4+x^4}\le \frac{a^2}{n^2}$$

and the Weierestrass M-Test guarantees that the series converges uniformly on any bounded interval.

Answer 2

If we have:

$$S_m(x) = \sum_{n=1}^m \frac{n^2x^2}{n^4+x^4},S(x) = \sum_{n=1}^\infty \frac{n^2x^2}{n^4+x^4}$$

then

$$\sup_\mathbb{R}\vert S_m(x)-S(x)\vert\ge\vert S_m(m+1)-S(m+1)\vert=\sum_{n=m+1}^\infty \frac{n^2(m+1)^2}{n^4+(m+1)^4}\ge\frac{(m+1)^2(m+1)^2}{(m+1)^4+(m+1)^4}=\frac{1}{2}$$

so $\vert\vert S_m-S\vert\vert_\infty$ does not converge to $0$, and hence our series does not converge uniformly on $\mathbb{R}$.

As noted in an earlier post, it does converge uniformly on bounded subsets of $\mathbb{R}$.

Answer 3

This is not an answer but it is too long for a comment. It is just a curiosity.

I worked this summation long time ago and identified that $$S(x) = \sum_{n=1}^\infty \frac{n^2x^2}{n^4+x^4}=\frac{\sqrt{2}\pi x}{4}\,\frac{\sin \left(\sqrt{2} \pi x\right)-\sinh \left(\sqrt{2} \pi x\right)}{\cos \left(\sqrt{2} \pi x\right)-\cosh \left(\sqrt{2} \pi x\right)}$$ which is defined for any value of $x$.