Show $\sum^\infty_{n=1}(\frac{x}{n^{0.6}(1+nx^2)})$ converges uniformly on $\mathbb{R}$ $\sum^\infty_{n=1}\frac{x}{n^{0.6}(1+nx^2)}$ converges uniformly on $\mathbb{R}$
Is $x\rightarrow\sum^\infty_{n=1}(\frac{x}{n^{0.6}(1+nx^2)})$ continuous at all points of $\mathbb{R}$?
I'm stuck on the first step, I feel I want to try and bound the expression inside the sum above. I've tried that and I'm not sure how.
I'm not that confident it is right either, because $x\sum^\infty_{n=1}(\frac{1}{n^{0.6}})$ doesn't converge.
IIRC $\frac{1}{x^p}$ converges for |p|>1, needs investigation for |p|=1 and doesn't converge for |p|<1 - from the ratio test.
 A: Let
$$
f_n(x):=\frac{x}{n^{0.6}(1+nx^2)}\qquad f(x):=\sum_{n\geq 1}f_n(x)
$$


*

*Check that this series of functions converges pointwise over $\mathbb{R}$ (this means that for each $x\in\mathbb{R}$ fixed, the series converges: you'll have to treat the case $x=0$ which is trivial separately; for the case $n\neq 0$, note that $|f_n(x)|\leq \frac{1}{xn^{1.6}}$)

*Check that $\sup_{x\in\mathbb{R}} |f_n(x)|=f_n(1/\sqrt{n})=\frac{1}{2n^{1.1}}$.

*Deduce from the latter that 
$$\sup_{x\in\mathbb{R}} \big| f(x)-\sum_{n=1}^kf_n(x)\big|\leq \sum_{n\geq k+1} \frac{1}{2n^{1.1}}\longrightarrow 0$$
as $k\rightarrow +\infty$ which is precisely uniform convergence of this series of functions over $\mathbb{R}$ (we say that a sequence of functions $g_k$ converges to $g$ uniformly over a set $S$ if $\sup_{x\in S}|g_k(x)-g(x)|\longrightarrow 0$ as $k$ tends to $+\infty$; we say a series of functions converges uniformly if the sequence of partial sums converges uniformly).


*

*Now you have probably seen that the uniform limit of a sequence (or series) of continuous functions is continuous. Otherwise, this is a classical $\epsilon/3$ exercise which is most likely a theorem in your book.

