Exercise Show that the series $\displaystyle\sum_{n=1}^\infty \frac{1}{x^3+n^3}$ converges uniformly on $[0, \infty)$
To approach this problem, I suspect it would be helpful to do some algebra to show that this is less than a power series. Since power series are uniformly convergent on $(-1,1)$ it would only seem to make sense to somehow turn the given summation into a power series of some sort or to relate it to a power series.
Could anyone provide me with some insight as to starting this problem? Or would it follow instead by the Weierstrass M-Test that this summation is uniformly convergent? I'm thinking the M-Test
Attempt
We consider $\displaystyle\sum_{n=1}^\infty \frac{1}{x^3+n^3}$. We note that for each $n \in \mathbb{N}$ and $x \in [0,\infty)$, we have
$$f_n(x) : = \frac{1}{x^3+n^3} \leq \frac{1}{n^3}$$
Here, $\{\frac{1}{n^3}\}_{n=1}^\infty$ is our collection of constants and for each $M_k \in \{\frac{1}{n^3}\}_{n=1}^\infty$, we have $f_n(x) \leq M_n$.
We note finally that
$$\displaystyle\sum_{k=1}^\infty M_k = \displaystyle\sum_{n=1}^\infty \frac{1}{n^3} < \infty$$
And thus, by the Weierstrass M-Test, it must follow that $\displaystyle\sum_{n=1}^\infty \frac{1}{x^3+n^3}$ converges uniformly on $[0,\infty)$.
Does this seem reasonable? Does the M-Test apply here?