Why is $\lim\limits_{n\to \infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ not equal to $1$?

Question

So I saw the limit $\lim\limits_{n\to \infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ here the other day:

Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$

and when I saw it, I right away thought the answer is $1$ because I thought $\lim\limits_{n\to \infty} \sum_{k=0}^n \frac{n^k}{k!} = \lim\limits_{n\to \infty} e^n$ given that $e^x = \lim\limits_{n\to \infty} \sum_{k=0}^n \frac{x^k}{k!}$ and so the result would be $\lim\limits_{n\to \infty} e^{-n}e^n = 1$ but the result is $\frac{1}{2}$, found using methods that I'm not familiar with.

Could someone please explain why my method is wrong?

Thank you so much in advance!

Answer 1

What you're doing is taking the identity $$ \lim_{n\to\infty}\sum_{k=0}^n\frac{x^k}{k!}=e^x\tag1 $$ and plugging in $x=n$ to obtain the (false) statement $$ \lim_{n\to\infty}\sum_{k=0}^n\frac{n^k}{k!}=e^n.\tag2 $$ Why is (2) false? Setting $x=n$ in (1) is illegal because the $n$ in (1) is busy being used as the label for the $n$th term in your sequence; plugging $x=n$ confuses $x$ with $n$ and changes the nature of the expression you're studying. To see why (2) makes no sense, notice that the LHS of (2) should no longer depend on $n$ when you've passed to the limit, so the RHS should not depend on $n$ either. For more examples of what can go wrong, try setting $x=n$ in the following identities, which are valid for all $x$: $$ \lim_{n\to\infty}\frac xn=0\tag3 $$ and $$ \lim_{n\to\infty}\left(1+\frac xn\right)^n=e^{x}\tag4 $$

Answer 2

Your argument is wrong because it doesn't make sense to talk about $\lim_{n \to \infty} \sum_{k=0}^n \frac{n^k}{k!}$. That limit doesn't exist (because it's infinite). You can talk about the sum's asymptotic behavior as compared to $e^n$, but that's not the same thing and it's not obvious (and if your result is correct, not true) that the sum asymptotically approaches $e^n$ as $n \to \infty$.

Answer 3

If the inner sum were added for k = 0 to infinity, you would be right. But it is only added for k = 0 to n. It seems that if you add the first (n+1) terms of the sum up to n, you get about half of the complete infinite sum (which you would need to prove), that is about $e^n / 2$.

Answer 4

Somebody gotta do this via CLT:

Let $X_n$ be a random variable, having Poisson distribution with parameter $n$, that is, $\mathbb{P}(X_n=k)=e^{-n}n^k/k!$, for every $k\geq 0$ integer. Now, $$ e^{-n}\sum_{k=0}^n \frac{n^k}{k!} = \mathbb{P}({\rm Pois}(n) \leq n). $$ Now, let $Y_1,\dots,Y_n$ be random variables with ${\rm Pois}(1)$ distribution, and thus, the given probability is nothing but $$ \mathbb{P}(Y_1+\cdots+Y_n \leq n) = \mathbb{P}\left(\frac{Y_1+\cdots+Y_n-n}{\sqrt{n}}\leq 0\right). $$ Now, since the central limit theorem tells us that $$ \frac{Y_1+\cdots+Y_n-n}{\sqrt{n}}\to N(0,1) $$ in distribution, we have therefore: $$ \lim_{n\to\infty}e^{-n}\sum_{k=0}^n \frac{n^k}{k!} = \lim_{n\to\infty} \mathbb{P}\left(\frac{Y_1+\cdots+Y_n-n}{\sqrt{n}}\leq 0\right) = \mathbb{P}(N(0,1)\leq 0), $$ which is $1/2$.