Uniform convergence of $\sum_{n=1}^{\infty}\frac{nx}{1+n^5x^2}$ verfication

I would like to know if my proof is correct:

I want to show that the series:

$$a_n=\sum_{n=1}^{\infty}\frac{nx}{1+n^5x^2}$$

uniformly converges for $|x|\lt \infty$

I looked at $a_n\lt b_n=\sum_{n=1}^{\infty}\frac{nx}{n^5x^2}=\sum_{n=1}^{\infty}\frac{1}{n^4x}$

I used d'Alambert therorem: $\frac{a_{n+1}}{a_n}=\frac{n^4x}{(n+1)^4x}=(\frac{n}{n+1})^4\lt1$

so $a_n$ also converges for every x.

I'm not sure if showing that the series converges for for every x is enough since my book is not clear enough about it.

by the section in the book I think I was expected to use weierstrass' uniform convergence theorem but I can't find a larger series that is independent from x.

• I guess, the $a_n$ and $b_n$ should be defined without the sum, right? – Michele Jul 12 '17 at 14:23

Using the inequality $\frac{2|ab|}{a^2 + b^2}\leq 1$ for every $a,b\in\mathbb{R}$ such that $(a,b)\neq (0,0)$, you get $$\left| \frac{nx}{1+n^5 x^2} \right| = \frac{1}{2 n^{3/2}} \cdot \frac{2 n^{5/2} |x|}{1+ n^5 x^2} \leq \frac{1}{2 n^{3/2}}, \qquad \forall x\in\mathbb{R},$$ hence the uniform convergence follows from the M-test.
• The series $\sum 1/(n^4 x)$ is defined (and convergent) for every $x\neq 0$. Unfortunately, it is not uniformly convergent in $\mathbb{R}\setminus\{0\}$. – Rigel Jul 12 '17 at 14:53
• Yes, basically it's the same. In the first inequality, the equality sign (i.e. the max) is achieved when $|a|=|b|$, i.e. for $n^{5/2} |x| = 1$. Anyway, you obtain the same estimate. – Rigel Jul 12 '17 at 16:32