Uniform convergence of $f(x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}x}{1+n^4x^2}$ on $\Bbb R$

Question

Let $f(x) = \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}x}{1+n^4x^2}$ for $x \in \Bbb R$. I have to study the uniform convergence of the series.

So I did like this.

$$\frac{(-1)^{n-1}x}{1+n^4x^2}\le \frac1{n^4}<\frac1{n^2}$$

Because $\frac1{n^2}$ converges $\implies$our functional series converges uniformly. Is this correct?

Answer 1

$$\bigg|\frac{(-1)^{n-1}x}{1+n^4x^2}\bigg|=\frac{|x|}{1+n^4x^2}=\frac{1}{2n^2}\cdot\frac{2n^2|x|}{1+n^4x^2}\leq\frac{1}{2n^2}\text{ for any }x\in\Bbb R.$$

As $$\frac{2|a||b|}{a^2+b^2}\leq 1\text{ for }(a,b)\not=(0,0).$$

So, apply Weierstrass $M$-test as $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}<\infty$.