Limit of sum function for infinite series $f(x)=\sum_{n=1}^{\infty}\frac{1}{n^6+x^4}$ as $x\rightarrow\infty$ As the title states I would like to determine the limit of $f(x)=\sum_{n=1}^{\infty}\frac{1}{n^6+x^4}$ as $x\rightarrow\infty$. My gut instinct here tells me that the limit should be 0 as each of the terms would go to 0, however I am having difficulty finding any formal reason as to why:
$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}f_n(x)=\sum_{n=1}^{\infty}\lim_{x\rightarrow\infty}f_n(x)$
If it were a finite sum then this would be easy enough, however the fact that I'm working with an infinite sum is causing me some trouble, I'd deeply appreciate any help!
 A: The result follows directly by Dominated Convergence. But here's an elementary argument. Fix $\varepsilon>0$. Choose $m$ such that $\sum_{n>m}\frac1{n^6}<\frac\varepsilon2$. If $x>(2m/\varepsilon)^{1/4}$, then
$$
\sum_n\frac1{n^6+x^4}\leq\frac\varepsilon2+\sum_{n=1}^m\frac1{n^6+x^4}
\leq\frac\varepsilon2+\sum_{n=1}^m\frac1{x^4}
=\frac\varepsilon2+\frac m{x^4}<\varepsilon.
$$
A: $$\frac{1}{n^6+x^4} \leqslant \frac{1}{n^6}$$
So you can get any property you would like from uniformly convergence:
General theorem about interchange limit and summation for uniformly convergence series holds also for any point $x_0$, finite or not("ε/3 trick"), which is density point for $f$ domain, so you can use $$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}f_n(x)=\sum_{n=1}^{\infty}\lim_{x\rightarrow\infty}f_n(x)$$
A: The limit is $0$ because, for any large $N\in\Bbb N$,$$f(N)=\sum_{n=1}^N\frac{1}{n^6+N^4}+\sum_{n=N+1}^\infty\frac{1}{n^6+N^4}\le\underbrace{\sum_{n=1}^N\frac{1}{N^4}}_{N^{-3}}+\underbrace{\sum_{n=N+1}^\infty\frac{1}{n^6}}_{\sim\tfrac15N^{-5}}\in N^{-3}+O(N^{-5}).$$Since $f$ decreases with increasing $N$, $f(x)\le f(\lfloor x\rfloor)$ for all $x\ge0$.
