Convergence of a series of functions in terms of a parameter Consider the sequence of functions $\{f_{n,\alpha}\}_{n\in \mathbb{N}}$ defined by $$f_{n,\alpha}(x)=\frac{|x|^n}{x^{2n}+n^{\alpha}}$$ where $x\in \mathbb{R}$ and $\alpha>0$. I have to study the intervals of pointwise, uniform and normal convergence of $\sum\limits_{n=1}^{\infty}f_{n,\alpha}$ in terms of $\alpha$. 
It's pretty easy to see by comparison with $|x|^n$ and $\frac{1}{|x|^n}$ that the series converges pointwise in $\mathbb{R}\setminus \{\pm 1\}$ for every $\alpha>0$ and since $f_{n,\alpha}(1)\sim \frac{1}{n^{\alpha}}$, the series converges pointwise in all $\mathbb{R}$ iff $\alpha>1$.
Also, studying $f'_{n,\alpha}(x) = \frac{n|x|^{n-1} (n^{\alpha}-x^{2n})}{(x^{2n}+n^{\alpha})^2}$ we get that $\sup\limits_{x\in \mathbb{R}} f_{n,\alpha}(x)=\frac{1}{2n^{\alpha/2}}$ attained at $x=n^{\frac{\alpha}{2n}}$, thus we have normal convergence in all $\mathbb{R}$ iff $\alpha>2$.
Now the problem is uniform convergence. Clearly if $\alpha>2$ the convergence is uniform in all $\mathbb{R}$, since the convergence is normal. Also, we already know that the convergence is uniform in all intervals of the form $[-a,a]$ with $0\leq a<1$ and $(-\infty,b]\cup [b,+\infty]$ with $b>1$ for every $\alpha>0$. We can extend uniform convergence to $[-1,1]$ iff $\alpha>1$, since $$\sup\limits_{x\in [-1,1]} f_{n,\alpha}(x)=f_{n,\alpha}(1)=\frac{1}{1+n^{\alpha}}$$ so by Weierstrass criterion we have uniform convergence if $\alpha>1$, and if $\alpha\leq 1$ the convergence can't be uniform in $[-1,1]$ since each $f_{n,\alpha}(x)$ is continuous but the sum $f(x)$ would go to infinity in $x=\pm 1$.
However, what about the uniform convergence in $[1,+\infty)$ when $1<\alpha\leq 2$? My problem is the right neighbourhood of $1$, since for large enough $n$ we have that $n^{\frac{\alpha}{2n}}\in [1,1+\delta]$ thus $\sup\limits_{x\in [1,1+\delta]} f_{n,\alpha}(x)=\frac{1}{2n^{\alpha/2}}$ and Weierstrass criterion for uniform convergence is inconclusive.
 A: Fix $\alpha\in(1,2]$, and the domain $[1,\infty)$.
First, you see that there is no normal convergence, as $$\lVert f_{n,\alpha}\rVert_\infty = f_{n,\alpha}(n^{\alpha/(2n)} = \frac{1}{2n^{\alpha/2}}$$ and thus $\sum_{n=1}^\infty \lVert f_{n,\alpha}\rVert_\infty$ diverges.
But normal convergence is stronger than uniform convergence, so this does not preclude uniform convergence. However, the following does:
$$
\sup_{x\geq 1} \left\lvert f(x)-\sum_{n=1}^N f_{n,\alpha}(x)\right\rvert
= \sup_{x\geq 1} \sum_{n=N+1}^\infty f_{n,\alpha}(x) \geq \sup_{x\geq 1} \sum_{n=N+1}^{2N}f_{n,\alpha}(x)\geq \sum_{n=N+1}^{2N}f_{n,\alpha}(N^{\alpha/(2N)})
$$
but
$$\begin{align}
\sum_{n=N+1}^{2N}f_{n,\alpha}(N^{\alpha/(2N)})
&= \sum_{n=N+1}^{2N}\frac{N^{\alpha/2}}{N^{\alpha}+n^\alpha}
= \frac{1}{N^{\alpha/2}}\sum_{n=N+1}^{2N}\frac{1}{1+\left(\frac{n}{N}\right)^\alpha}
\geq \frac{1}{N^{\alpha/2}}\sum_{n=N+1}^{2N}\frac{1}{1+2^\alpha}\\
&= N^{1-\alpha/2}\frac{1}{1+2^\alpha}
\end{align}$$
which does not go to $0$ as $N\to\infty$.

Note that the issue is indeed on the right-neighborhood of $1$, as 
$$
n^{\frac{\alpha}{2n}} = 1+\frac{\alpha\log n}{2n} + o\left(\frac{1}{n}\right)
$$
