Uniform convergence of $\sum_{n=1}^{\infty} \frac{1}{1+n^3x}$ I need to find where in $\mathbb{R}$ the following series converges absolutely and where it converges uniformly:
$$f(x) = \sum_{n=1}^{\infty} \frac{1}{1+n^3x}$$
I know by the M-test that both conditions are satisfied when $x>0$ simply from the bound $|f_n(x)| \leq \frac{1}{x} \frac{1}{n^3}$. For $x = 0$, neither are possible, as the series diverges there.
But for $x < 0$, I do not how to find a bound on the $f_n(x)$. I've tried reverse triangle inequality and expanding $f_n$ in series, but this does not help, and also I do not know where the series even converges for these values of $x$. Is M-test still the correct approach?
 A: The Weierstrass M-test does imply uniform convergence on any interval $[a,\infty)$ where $a > 0$ since
$$|f_n(x)| \leqslant \frac{1}{1+n^3a},$$
and the series with terms given by the RHS is convergent.
However, the convergence is not uniform for all $x  >0$  since $f_n(x) =  \frac{1}{1+ n^3x}$ does not converge uniformly to $0$ on $(0,\infty)$. This is evident because
$$\sup_{x \in (0,\infty)}|f_n(x)| = f_n(0) =1 \underset{n \to \infty}\longrightarrow1 \neq 0$$
Alternatively, we see that
$$\sup_{x \in (0,\infty)}\left|\sum_{k=n+1}^\infty \frac{1}{1+k^3x}\right|= \sup_{x \in (0,\infty)}\sum_{k=n+1}^\infty \frac{1}{1+k^3x}\geqslant \sup_{x \in (0,\infty)}\sum_{k=n+1}^{2n} \frac{1}{1+k^3x} \\\geqslant n \cdot \frac{1}{1+(2n^3)\cdot n^{-1/3}}= \frac{n}{3}\underset{n \to \infty}\longrightarrow +\infty \neq 0$$
For $x \in (-\infty,-a]$ where $-a < -1$, we can again apply the Weierstrass M-test to find uniform convergence, since $x \leqslant -a < -1$ and $1+n^3x \leqslant 1-n^3a < 1 - n^3$. This implies that $|1+n^3x| \geqslant n^3a-1 >0 $
$$|f_n(x)| = \frac{1}{|1+n^3x|} \leqslant \frac{1}{n^3a -1} = \mathcal{O}(n^{-3})$$
Try to finish by considering when $x \in [-1,0)$.
