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.
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 $\infty$.
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).
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 $$ \lim_{n \to \infty} \left(\left( e^{-n} \sum_{i=1}^n \frac{n^i}{i!} \right) + e^{-n} \right) = \lim_{n \to \infty} e^{-n} \sum_{i=\color{red}{0}}^n \frac{n^i}{i!}. $$
A probabilistic approach. Let $X_n\sim \text{Poi}(n)$ be a Poisson random variable with rate $\lambda=n$. By the central limit theorem, it follows that $$ Z_n=\frac{X_n-n}{\sqrt{n}}\stackrel{d}{\to} Z\sim N(0,1). $$ Your limit equals $$ \lim_{n\to \infty} P(X_n\leq n)=\lim_{n\to \infty}P(Z_n\leq 0)=P(Z\leq 0)=1/2. $$
As TheSilverDoe commented, this is not immediate.
In fact $$\sum_{i=1}^{n}\frac{n^i}{i!}=\frac{e^n \Gamma (n+1,n)}{n \Gamma (n)}-1$$ where appears the incomplete gamma function. So $$e^{-n}\sum_{i=1}^{n}\frac{n^i}{i!}=\frac{ \Gamma (n+1,n)}{n \Gamma (n)}-e^{-n}$$
The main problem with the reasoning is that the series for $e^n=\sum_{i=0}^\infty {n^i \over i!}$ gets truncated at a point (after $i=n$), where the remaining series still makes a non-trivial contribution to $e^n$. The first missing part is $n^{n+1} \over (n+1)!$, which is still $>> 1$.
That means there is no $n$ where the remaining contribution of the series is small and one could say "the finite series considered in the problem is $e^n-\epsilon$". That the limit in the problem statement is $1\over2$ is not at all trivial or intuitive, it's quite strange that the limit exists at all.
Note that nobody in any other answer (as of now) could actually prove it, where did you get to "see" that this is indeed the case?
