Prove that $\sum_{n=1}^{\infty}\frac{x^2}{1+n^3x^4}$ converges uniformly for $x\in \mathbb{R}$.

Question

Prove that $$\sum_{n=1}^{\infty}\frac{x^2}{1+n^3x^4}$$ converges uniformly for $x\in \mathbb{R}$.

My try: for $x=0$, of course it converges. for $x\neq 0$ , then $\frac{x^2}{1+n^3x^4}\le \frac{1}{n^3x^2}$ hence converges.

I am uncertain for my try. Please verify it.

Answer 1

Using standard real-analysis methods, you can prove that the maximum of $\frac{x^2}{1+n^3x^4}$ is attained when $x=\pm n^{-\frac34}$. That maximum is $\frac12n^{-\frac32}$. Therefore, by the Weierstrass $M$-test and since the series $\sum_{n=1}^\infty\frac12n^{-\frac32}$ converges, the convergence is uniform in $\mathbb R$.