Assume we have some series of functions defined for any $x\geq0$:
$$S(x)=\sum_{n=1}^{\infty} \frac{3x+n}{x+n^{3}}$$
Let's assume uniform convergence. Meaning
$$\forall{\epsilon\geq0}, \exists N s.t. \forall{n \geq N}, \forall{p}, \forall{x\geq0}$$ $$ \lvert \frac{3x+(n+1)}{x+(n+1)^{3}} + \frac{3x+(n+2)}{x+(n+2)^{3}} + \cdots + \frac{3x+(n+p)}{x+(n+p)^{3}} \rvert \leq \epsilon $$
Let's pick a specific $n_0 \geq N$ and $p=1$:
$$ \lvert \frac{3x+(n_0)}{x+(n_0)^{3}} \rvert = \frac{3x+(n_0)}{x+(n_0)^{3}} \leq \epsilon $$
We know this is false, because $\lim_{x\to\infty}=3$ which obviously greater than many $\epsilon$.
However, this is the part I don't really understand... If instead of infinity, we approach some number $M\lt \infty$ then the series is uniformly convergent. Why?
What's to stop us from picking $M+1$ and continuing the series?
Just asserting that a series of functions is uniformly convergent is meaningless. What has a meaning is to assert that a series of functions converges uniformly (or not) on some given set. Your series does not converge uniformly on $[0,\infty)$, but it does converge uniformly on any bounded subset of $[0,\infty)$. There is no contradiction here.