Show that $\sum_{n=1}^\infty \ \frac{x}{n(1+nx^2)}$ converges uniformly on $R$.

Question

I need help with this problem:

Show that $$\sum_{n=1}^\infty \ \frac{x}{n(1+nx^2)}$$ converges uniformly on $R$.

So, I tried using the Weierstrass $M$-test. This is what I did:

$f_n(x)=\frac{x}{n(1+nx^2)}$ so by the $M$-test $|f_n(x)|\leq M_n$, so $\left|\frac{x}{n(1+nx^2)}\right|\leq \left|\frac{x}{n(nx^2)}\right|=\left|\frac{x}{n^2x^2}\right|\leq \frac{1}{n^2}$, and $\frac{1}{n^2}$ converges, so the series converges uniformly on $R$.

Is what I'm doing right? Because the assistant teacher said that we can't use the $M$-test here, why? If I'm wrong, how can I solve this correctly?

Answer 1

Note that $1 + nx^2 \geqslant 2\sqrt{n}|x|$ since $(1 - \sqrt{n}|x|)^2 \geqslant 0 $ and

$$ \left|\frac{x}{n(1+nx^2)}\right| = \frac{|x|}{n(1+nx^2)}\leqslant \frac{|x|}{n}\frac{1}{2|x|\sqrt{n}} = \frac{1}{2n^{3/2}}$$

By the Weierstrass M-test the series is uniformly convergent for $x \in \mathbb{R}$ since $\sum_{n \geqslant 1} n^{-3/2}$ is a convergent p-series.

Answer 2

Let $f_n(x)$ be the $n$th summand. A little calculus shows the maximum value of $f_n$ over $[0,\infty)$ occurs at $x= 1/\sqrt {n},$ giving a value of $\dfrac{1}{2n^{3/2}}.$ Because $f_n$ is odd, we have $\sup_{\mathbb R}|f_n|\le \dfrac{1}{2n^{3/2}}.$ By Weierstrass M, the series converges uniformly on $\mathbb R.$