Proving partial sums $A_n = o(|z_k|^\rho)$, where $|z_k|\to\infty$ is increasing The following is a question in Lang's Complex Analysis, fourth edition, Chapter XIII $\S 4$, problem $5a$:

Let $\{ a_k \} , \{ z_k \}$ be sequences of complex numbers such that $|z_k| \to \infty$ and $|z_k| \le |z_{k+1}|$.  Let $\rho > 0$ such that $$\sum_{k=1}^\infty \frac{|a_k|}{|z_k|^\rho} < \infty$$  Let $A_n$ denote the partial sums of $|a_k|$.  Prove that $A_n = o(|z_n|^\rho)$.

I am having difficulty proving the little-oh part of it.  It's very easy to prove $A_n = O(|z_n|^\rho)$, since
$$\lim_{n\to\infty} \frac{A_n}{|z_n|^\rho} \le \sum_{k=1}^\infty \frac{|a_k|}{|z_k|^\rho} < \infty$$
However, it's not so clear why the limit should go to $0$.  My attempt so far has been to assume that the conclusion does not hold so that $A_n \ge \epsilon |z_n|^\rho$ for infinitely many $n$.  We can rewrite the sum as:
$$\sum_{k=1}^\infty \frac{|a_k|}{|z_k|^\rho} = \sum_{k=1}^\infty \left( \frac{1}{|z_k|^\rho} - \frac{1}{|z_{k+1}|^\rho} \right) A_k \approx \epsilon \sum_{k} 1 - \left( \frac{|z_k|}{|z_{k+1}|}\right)^\rho$$
The last sum seems like it might diverge (although I have my own doubts) but I'm stuck at this point.
 A: Let $c_k=|a_k|/|z_k|^\rho \text { and } b_k=|z_k|^\rho$
Then
$$\lim_{n\to\infty} \frac{1}{b_n}\sum_{k=1}^n c_kb_k=0$$
Why the $\lim_{n\to\infty} \frac{1}{b_n}\sum_{k=1}^n c_kb_k=0$:
Let $a_n$ be a sequence of complex numbers such that $\sum a_n$ converges. Let $b_n$ be a sequence of real numbers which is increasing, i.e., $b_n \le b_{n+1}$ for all $n$, and $b_n\rightarrow\infty$ as $n\rightarrow\infty$
Given $\epsilon>0$, if we select a positive integer $n_0$ such that for all $m>n_0$ we have $\left|\sum_{n=n_0+1}^m a_n\right|<\epsilon$ and $b_{n_0}\ge0$. Then for $N>n_0$ splitting the sum we obtain:
$$\left|\frac{1}{b_N}\sum_{n=1}^N a_nb_n\right|\le\left|\frac{1}{b_N}\sum_{n=1}^{n_0} a_nb_n\right|+\left|\frac{1}{b_N}\sum_{n=n_0+1}^N a_nb_n\right|$$
The first sum will be $\le\epsilon$ for all large N. For the second sum we use summation by parts to obtain, after some elementary computations,
$$\sum_{n=n_0+1}^N a_nb_n=b_N(A_N-A_{n_0})-\sum_{k=n_0+1}^{N-1} (A_k-A_{n_0})(b_{k+1}-b_k)$$
where $A_n=\sum_{k=1}^n a_k$ are the partial sums. Therefore by the triangle inequality the fact that $\left|A_k-A_{n_0}\right|<\epsilon$ for all $k\ge n_0$ and that $b_k$ increases we get:
$$\left|\sum_{n=n_0+1}^N a_nb_n\right|\le\left| b_N\right|\epsilon+\epsilon(b_N-b_{n_0})$$
Hence for all large $N$ we have:
$$\left|\frac{1}{b_N}\sum_{n=n_0+1}^N a_nb_n\right|\le 3\epsilon$$
which concludes our proof.
However if we only assume that the partial sums of $\sum a_n$ are bounded, we cannot conclude that the limit is $0$. 
