Suppose $(\alpha_n)_{n=0}^\infty$ is a sequence in $\mathbb{C}$. If $\alpha_n\rightarrow \alpha$, why is it true (for $\epsilon>0$ given) that if I choose $n_1>n_0$ such that
\begin{equation} \frac{1}{n+1}\sum_{k=0}^{n_0}|\alpha - \alpha_k|<\epsilon/2 \end{equation}
for all $n>n_1$?
I don't understand why this sum can be bounded by $\epsilon$, specially because it involves the first terms in the sequence. What it is true is that for $\epsilon>0$, I can choose $n_0\in\mathbb{N}$ such that $|\alpha - \alpha_k|<\epsilon/2$ for all $k>n_0$. Thus, one can bound easily the sum
\begin{equation} \sum_{k=n_0+1}^{n_1}|\alpha - \alpha_k|, \end{equation}
but this does not help to bound the terms $|\alpha-\alpha_0|, |\alpha-\alpha_1|,\dots,|\alpha - \alpha_{n_0}|$.