asymptotics of sum of $\ln$ Let $(z_n)_{n\in\mathbb N}$ and $(u_n)_{n\in\mathbb N}$ be sequences of complex numbers such that: $\lim_{n\to+\infty}|z_n|=+\infty$, $\lim_{n\to+\infty}|u_n|=+\infty$, $|u_n|=o(|z_n|)$ and $(|u_n|)_{n\in\mathbb N}$ is non-decreasing. I want to prove that
$$\sum_{k=0}^{n-1}\ln|z_n-u_k|\underset{n\to+\infty}{\sim} n\ln|z_n|.$$
I did not manage to prove that. Any proof or hint will be welcome. Thanks in advance.
 A: Since $|u_n|=o(|z_n|)$ for $n\to \infty$ and since $|u_k|$ is non-decreasing, we have for sufficiently large $n$ and $k\le n$ 
$$\left|\frac{u_k}{z_n}\right|\le 1/2 \tag1$$
With $n\ge k$ chosen so that $(1)$ holds, application of the triangle inequality reveals
$$\frac12\le 1-\left|\frac{u_k}{z_n}\right| \le \left|1-\frac{u_k}{z_n}\right|\le 1+\left|\frac{u_k}{z_n}\right|\le \frac32\tag 2$$

Now, write the summand, $\log\left(\left|z_n-u_k\right|\right)$, as
$$\log(|z_n-u_k|)=\log(|z_n|)+\log\left(\left|1-\frac{u_k}{z_n}\right|\right)\tag 3$$
we find that
$$\begin{align}
\left|\sum_{k=0}^{n-1}\log\left(\left|z_n-u_k\right|\right)\right|&=n\log(|z_n|)+\sum_{k=0}^{n-1}\left|\log\left(\left|1-\frac{u_k}{z_n}\right|\right)\right|\\\\
&\le n\log(|z_n|)+n\log(2)\\\\
&\sim n\log(|z_n|)
\end{align}$$
as was to be shown!
A: I had misread the question. For any $\epsilon > 0,$ there exists an $N$ such that for all $n\geq N,$ $|u_n/z_n| < \epsilon,$ and the same is true for all $k\leq n,$ by monotonicity. So, 
$$\left|\sum_{k=0}^{n-1} \log |z_n - u_k| - n \log|z_n|\right| =
\left|\sum_{k=0}^{n-1}\log(1-\frac{u_k}{z_n}) \right|\leq n\epsilon.$$
Now, divide through by $n\log |z_n|$ to get 
$$\frac{\left|\sum_{k=0}^{n-1} \log |z_n - u_k|\right|}{n \log|z_n|}-1\leq \epsilon/|z_n|\leq \epsilon,$$for $n>N,$ and also large enough so that $|z_n|>1.$
