sequences and series - Why does $limlimits_{ntoinfty} e^{-n}sum_{i=1}^{n}frac{n^i}{i!} = frac{1}{2}$ and not 1?

The limit
$$\lim_{n\to\infty} e^{-n}\sum_{i=1}^{n}\frac{n^i}{i!}$$
can be seen to be $\frac{1}{2}$, yet isn't the sum in this expression just going to be $\lim\limits_{n\to\infty}e^{n}$, making the limit 1?

I'm having trouble wrapping my head around why this isn't the case. Any help would be much appreciated.

Answer

1. The problem with your reasoning is that the two terms, $e^{-n}$ and $\sum_{i=1}^n \frac{n^i}{i!}$, can't be analyzed separately. Notice that $e^{-n}$ approaches $0$, and the second term approaches $\infty$, so the limit of the product would be $\boldsymbol{0 \cdot \infty}$, an indeterminate form. A limit of the form $0 \cdot \infty$ might equal any real number, or might even equal $\infty$.





2. It may be instructive to consider a different expression where some $n$s are replaced by $m$. The following limit can be evaluated as you say (I also made the sum start from $i = 0$ for simplicity):
   $$
   \lim_{n \to \infty} e^{-\color{red}{m}} \sum_{i=0}^{\color{blue}{n}} \frac{{\color{red}{m}}^i}{i!} = 1,
   $$
   because it is the product of limits, $e^{-m} \cdot e^m = 1$. And if we instead take the limit as $m \to \infty$, then we get
   $$
   \lim_{m \to \infty} e^{-m} \sum_{i=0}^n \frac{m^i}{i!} = 0,
   $$
   because the exponential beats the polynomial, and goes to $0$. In your problem, essentially, $m$ and $n$ are both going to $\infty$ at the same time, so we might imagine that the two possible results ($0$ and $1$) are "competing"; we don't know which one will win (and it turns out that the result is $\frac12$, somewhere in the middle).




3. How can we show that your limit is $\frac12$? This is a difficult result; please take a look at this question for several proofs (thanks to TheSilverDoe for posting).

   
   
   
   

   In that question, the summation starts from $i=0$ instead of $i=1$. However, note that we can add $e^{-n}$ to your limit and it will not change (since $\lim_{n \to \infty} e^{-n} = 0$). So this gives
   $$
   \lim_{n \to \infty} \left(\left( e^{-n} \sum_{i=1}^n \frac{n^i}{i!} \right) + e^{-n} \right)
   = \lim_{n \to \infty} e^{-n} \sum_{i=\color{red}{0}}^n \frac{n^i}{i!}.
   $$