What happens to $n$ in the Limit of some function...

Question

Given that:

$$P(0,\alpha)\ \text{:=}\ 1-\alpha \\ P(1,\alpha)\ \text{:=}\ (1-\alpha ) \left(e^{\alpha }-1\right)\ \\ P(k,\alpha)\ \text{:=}\ (1-\alpha ) \sum _{j=0}^k \frac{e^{\alpha j} (-1)^{k-j} \left((\alpha j-j+k) (\alpha j)^{-j+k-1}\right)}{(k-j)!},\ k\geq 2$$

I'm interested in the following quantity, $S$, as $n\rightarrow\infty$

$$ S = \sum _{k=1}^j \left(1-\left(1-\frac{P(k,\alpha )}{\sum _{x=1}^k P(x,\alpha )}\right)^n\right). $$

So my attempt is to find the limit as $n\rightarrow\infty$, which is

$$ S = \lim_{n\rightarrow\infty}\left[\sum _{k=1}^j \left(1-\left(1-\frac{P(k,\alpha )}{\sum _{x=1}^k P(x,\alpha )}\right)^n\right) \right]. $$ and can be rewritten as: $$ S = \lim_{n\rightarrow\infty}\left[\sum _{k=1}^j \left(1-\left(1-\frac{n \cdot\frac{ P(k,\alpha )}{\sum _{x=1}^k P(x,\alpha )}}{n}\right)^n\right) \right], $$ which becomes... $$ S = \sum _{k=1}^j \left(1-\exp\left(-n \cdot\frac{ P(k,\alpha )}{\sum _{x=1}^k P(x,\alpha )}\right)\right). $$

The question I am asking is what is the interpretation of $n$ mean here? I.e., since I took $n \rightarrow \infty$ earlier?

Answer 1

There has been a mistake. If you take limits in some 'variable' $n$ and still have '$n$' in your final evaluation, then that should be a strong red flag.

Leaving out all the fluff, you've stated the following:

$$\tag{1}\lim_{n\to\infty}(1-nx/n)^n=e^{-nx}$$

Because you know that: $$\tag{2}\lim_{n\to\infty}(1-y/n)^n=e^{-y}$$

But $(1)$ is completely false. You cannot prove it using $(2)$, because there $y$ is simply some real number, and in $(1)$ we consider a non-constant function of $n$. What is true is that:

$$\tag{3}\lim_{n\to\infty}(1-\alpha_n/n)^n=e^{-\alpha}$$If $\lim_{n\to\infty}\alpha_n=\alpha$ and $(\alpha_n)_{n\in\Bbb N}$ is some convergent sequence of reals.

The correct evaluation of the limit in $(1)$ is: $$\lim_{n\to\infty}(1-x)^n=\begin{cases}0&0<x<2\\1&x=0\\+\infty&x<0\\\text{does not exist}&\text{otherwise}\end{cases}$$