Let $p\geq 1$ be an integer. If $\sum_{n=1}^\infty 1/|a_n|^{p+1}$ converges, show that $$\sum_{n=1}^\infty\left({1\over z-a_n}+{1\over a_n}+{z\over a_n^2}+\cdots+{z^{p-1}\over a_n^p}\right)$$ is meromorphic.
First note that the Taylor expansion of ${1\over z-a_n}$ is $$ -{1\over a_n}-{z\over a_n^2}-{z^2\over a_n^3}-\cdots.$$ It suffices to show the given series converges uniformly on each compact subset excluding the poles. Let $R>0$ and consider a compact subset $|z|\leq R$. Since $|a_n|\to\infty$ as $n\to\infty$, choose $n$ large so that $R/|a_n|<1/2$. Then \begin{align*} \left|{1\over z-a_n}+{1\over a_n}+{z\over a_n^2}+\cdots+{z^{p-1}\over a_n^p}\right| & = \left|-\sum_{k=p+1}^\infty{z^{k-1}\over a_n^k}\right|\\ &\leq\sum_{k=p+1}^\infty{|z|^{k-1}\over |a_n|^k}\\ &\leq{1\over |a_n|}\sum_{k=p+1}^\infty\left|{z\over a_n}\right|^{k-1}\\ &\leq {C\over|a_n|}\\ \end{align*} for some constant $C$. But this is not enough. I need to find a different estimate so that the final equality is $O(|a_n|^{-p-1})$. Could you help?
Note. It's possible that the problem has a typo.
You are on the right track. If $|z| \le R \le |a_n|/2$ then $$ \left|\frac{1}{z-a_n} + \frac{1}{a_n} + \frac{z}{a_n^2} + \cdots + \frac{z^{p-1}}{a_n^p} \right| \le \sum_{k=p+1}^\infty \left| \frac{z^{k-1}}{a_n^k}\right| = \left| \frac{z^{p}}{a_n^{p+1}} \right| \cdot \sum_{k=0}^\infty \left| \frac{z^{k}}{a_n^k}\right| \\ \le \frac{R^p}{|a_n|^{p+1}} \sum_{k=0}^\infty \left(\frac 12\right)^k = \frac{2R^p}{|a_n|^{p+1}} $$ and since $\sum_{n=1}^\infty 1/|a_n|^{p+1}$ converges, the conclusion follows.