(*) $\quad$ $\hspace{1mm} |\sum_{k=0}^{N}(\prod_{i=0}^{k-1}(\frac{n-i}{n}) - 1)\frac{z^{k}}{k!}| \hspace{1mm}$
Consider (*) for some bounded $N < n$.
I want to confirm that (*) is bounded by some $\epsilon > 0$ as $n \rightarrow \infty$.
I don´t know how to estimate this expression (*). I have observed that the product $\prod_{i=0}^{k-1}(\frac{n-i}{n})$ is close to one as $n \rightarrow \infty$. But how to claim that this is enough for the sum to be bounded by some $\epsilon$?
I mean according to the exercise we have: $\hspace{1mm} |\sum_{k=0}^{N}(\prod_{i=0}^{k-1}(\frac{n-i}{n}) - 1)\frac{z^{k}}{k!}| \leq \sum_{k=0}^{N}|\prod_{i=0}^{k-1}(\frac{n-i}{n}) - 1|\frac{|z|^{k}}{k!} < \dots < \epsilon$
But how to confirm that since $|\prod_{i=0}^{k-1}(\frac{n-i}{n}) - 1|$ tends to 1 in the limit.
I have also tried to find a pattern by looking at $k = 0,1,2,3,\dots$
$k=0 \Rightarrow (*) = 0$
$k=1 \Rightarrow (*) = 0$
$k=2 \Rightarrow (*) = \frac{1}{2!n}|z|^{2}$
$k=3 \Rightarrow (*) = \frac{3n-2}{3!n^{2}} |z|^{3}$
$k=4 \Rightarrow (*) = \frac{6n^{2}-11n-6}{4!n^{3}} |z|^{4}$
But I don´t recognize a pattern in the numerators $0, 0, 1, 3n-2, 6n^{2}-11n-6$.