How can one prove that $$f(z)=\sum_{n=1}^{\infty}\dfrac{z}{n(1+n|z|^{2})}$$ converges uniformly on $\mathbb{C}$ ?
My attempt was to let $z=x+iy$, then consider $$|\dfrac{z}{n(1+n|z|^{2})}|=\dfrac{\sqrt{x^{2}+y^{2}}}{\sqrt{(n^{2}x^{2}+n)^{2}+(n^{2}y)^{2}}}.$$
which looks like a mess. The main idea was to try to find an upper bound, $M_{k}$ so that I can apply the Weierstrass M-test. If I can show that the upper bound, $M_{k}$ is convergent (which is often simple by using root test etc.), I will be able to conclude that $f(z)$ converges uniformly.
Is there a simpler or more elegant way I can approach this question and if my approach is correct, can someone kindly guide me to find the required upper bound?
Finding an upper bound for $$ \left|\frac{z}{n(1+n|z|^{2})}\right | $$ uniformly for $z \in \Bbb C$ is equivalent to finding an upper bound for $$ \tag{*} \frac{r}{n(1+nr^2)} $$ uniformly for $r \ge 0$, therefore using the $z = x+iy$ representation is not needed and makes things only more complicated.
In order to estimate $(*)$, one can use the “inequality of arithmetic and geometric means”, as @Did demonstrated in above comment. But if you don't come up with that idea, just proceed systematically and determine (for fixed $n$) the maximum of the function $$ \phi(r) = \frac{r}{n(1+nr^2)} \, , \quad r \ge 0 \, . $$ It turns out that the (unique) maximum is at $r = \frac{1}{\sqrt n}$, and therefore $$ \left|\frac{z}{n(1+n|z|^{2})}\right | \le \frac{1}{2 n^{3/2}} \, . $$ for all $z \in \Bbb C$.