disk of convergence for complex-valued series Find the disk of convergence of $\displaystyle \sum_{k=0}^{\infty} \frac{(z+2)^k}{(k+2)^3 4^{k+1}}$, where $z \in \mathbb{C}$.
I tried applying the ratio test: $\lim_{k \to \infty} \left| \frac{(z+2)^{k+1}}{(k+3)^3 4^{k+2}} \cdot \frac{(k+2)^3 4^{k+1}}{(z+2)^{k}} \right| = \lim_{k \to \infty} \left| \frac{(z+2) \cdot (k+2)^3}{4 (k+3)^3} \right| = \left|\frac{z+2}{4} \right|.$
Do we just check where $\left|\frac{z+2}{4} \right|<1$ to get the radius of convergence? Not entirely sure where to go from here

I'd rather go directly with the $\;n\,-$ th root test (Cauchy-Hadamard formula) of the general term of the coefficients:
$$\sqrt[k]{|a_k|}=\sqrt[k]{\frac1{(k+2)^34^{k+1}}}=\frac1{4\sqrt[k]{4(k+2)^3}}\xrightarrow[k\to\infty]{}\frac14\;\implies R=4$$
and thus the interval of convergence ( around $\;-2\;$ , of course ) is the disk $\;|z+2|<4\;$ .
The way you went with the quotient test the answer is yes: just check when $\;\frac{|z+2|}4<1\;$
