Does the series $\sum_{n=1}^{\infty}\frac{n^2}{\sqrt{n!}}(x^n+x^{-n})$ converge for $x\in[\frac{1}{2},2]$?
I think that it does but not uniformly.
Since $$\left| \frac{n^2}{\sqrt{n!}}(x^n+x^{-n})\right|=\left| \frac{n^2}{\sqrt{n!}}\right||x^n+x^{-n}|\leq \left| \frac{n^2}{\sqrt{n!}}\right|\left|2^n+\frac{1}{2^n}\right|$$ and as $n\to \infty$ $$\lim_{n\to \infty}\frac{n^2}{\sqrt{n!}}\left(2^n+\frac{1}{2^n}\right)=\lim_{n\to \infty}\frac{2^n n^2}{\sqrt{n!}}=0 $$ and the series converges pointwise but I can't show that it doesn't converge unifromly. I am thinking about using the Cauchy criterion, but I dont know how I should start.
$$\left| \sum_{n=m+1}^{k} \frac{n^2}{\sqrt{n!}}(x^n+x^{-n})\right|\leq \sum_{n=m+1}^{k}\left| \frac{n^2}{\sqrt{n!}}(x^n+x^{-n})\right|\leq \sum_{n=m+1}^{k} \frac{n^2}{\sqrt{n!}}\left(2^n+\frac{1}{2^n}\right)$$ and I can't argue that for $n\to \infty$ the sum of the RHS diverges.
Any comment would be very helpful.