Show that function $f(x)=\sum_{n=1}^\infty \frac{nx}{1+n^4x^2}$ is uniformly convergent on $[a,\infty)$ I was asked to prove the series of function $$f(x)=\sum_{n=1}^\infty \frac{nx}{1+n^4x^2}$$ is uniformly convergent on $[a,\infty)$. (where $a>0$).
Here is what I've done:
I first consider the maximum value of $\frac{nx}{1+n^4x^2}$ by differentiating with respect to $x$, and I found out that the max occurs at $x=\frac{1}{n}$ and the value is $\frac{1}{2n}$.
So I have the following inequality: $$\sum_{n=1}^\infty \frac{nx}{1+n^4x^2} \le \sum_{n=1}^\infty \frac{1}{2n}$$
However, the series $\sum_{n=1}^\infty \frac{1}{2n}$ is divergent which means I cannot use the Weierstrass M-Test.
So I try to find another approach and arrive the following inequality:
$$\sum_{n=1}^\infty \frac{nx}{1+n^4x^2} \le \sum_{n=1}^\infty \frac{nx}{n^4x^2}=\sum_{n=1}^\infty \frac{1}{n^3x}\le \sum_{n=1}^\infty \frac{1}{n^3a}$$
I am not sure how to proceed next from here and also unsure if I am on the right track. Hope someone could give me a hint.
Thanks.
 A: You did good work, the only thing left to do is to notice that the series is uniformly convergent by Weierstrass $M$-test. The series $\sum_{n=1}^\infty \frac{1}{an^3}$ is indeed a convergent series of real numbers. 
Perhaps you got confused because the series of maximums you found is divergent, so that made you think you cannot use the $M$-test here. However, note that the maximum you found is at the point $x=\frac{1}{n}$, and for a sufficiently large $n$ this is already outside the interval $[a,\infty)$. So you really wouldn't be able to use $M$-test it if you wanted to test uniform convergence in $(0,\infty)$. But since you are interested only in the interval $[a,\infty)$ this is easier. 
A: You can try the following
$$\left|\frac{nx}{1+n^4x^2}\right|=\frac{n|x|}{1+n^4x^2}\le\frac{n|x|}{n^4x^2}=\frac1{n^3|x|}$$
and since $\;\sum\limits_{n=1}^\infty\cfrac1{n^3|x|}\;$ converges for any $\;0\neq x\in\Bbb R\;$ , you can apply Weierstrass M-test and we're done.
For $\;x=0\;$ the convergence of the original series is trivial, so your series converges for any real $\;x\;$ (and not only for positive ones)
A: In your second attempt, you are basically done by noting that the $p$-series $\displaystyle \sum_{n=1}^\infty \frac{1}{n^p}$ is convergent with $p>1$. 
To write it in a more "reasonable" way, observe that for any given $x\geq a$ and for each $n$, one has
$$
0\leq\frac{nx}{1+n^4x^2}\leq
\frac{nx}{n^4x^2} = \frac{1}{x}\cdot \frac{1}{n^3}\leq\frac{1}{a}\cdot\frac{1}{n^3}\;.\tag{1}
$$
Now applying the fact mentioned above and the direct comparison test, one concludes that your series converges for each $x\geq a$. The convergence is uniform on $[a,\infty)$ because the estimate (1) is independent of $x\in[a,\infty)$.
A: Let $f_m(x)=\sum_{n=1}^m nx/(1+n^4x^2).$ 
Given $r >0,$ take $m_r\in \Bbb N$ such that $\sum_{n=1+m_r}^{\infty}1/n^3<ar.$ Then for all $x\ge a$ and all $m\ge m_r$ we have $$|f(x)-f_m(x)|=\sum_{n=1+m}^{\infty}nx/(1+n^4x^2)\le$$ $$\le\sum_{n=1+m_r}^{\infty}nx/(1+n^4x^2)<$$ $$<\sum_{n=1+m_r}^{\infty}nx/(n^4x^2)=$$ $$=x^{-1}\sum_{n=1+m_r}^{\infty}1/n^3<$$ $$<x^{-1}ar\le a^{-1}ar=r.$$
