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!

"I thought that $\lim_{n \to \infty} \sum_{k=0}^n \frac{n^k}{k!} = e^n$". Huh? That makes absolutely no sense. There's an $n$ on the right hand side? — mathworker21, Nov 07 '19 at 21:57
@mathworker21 I meant $e^n$ as $n$ approaches infinity. Will fix it. — , Nov 07 '19 at 21:58
@GerryMyerson https://math.stackexchange.com/questions/160248/evaluating-lim-limits-n-to-infty-e-n-sum-limits-k-0n-fracnkk — , Nov 07 '19 at 22:00
@TBTD I have no idea what that is LOL. But I will try and read about it. — , Nov 07 '19 at 22:01
$\lim_{x \to \infty} x = \lim_{x \to \infty} x^2$ implies $\lim_{x \to \infty} \frac{x}{x^2} = 1$? the issue is that you need the common limit to be finite to deduce the limit of the ratio is $1$ — mathworker21, Nov 07 '19 at 22:01
We have for each $n \in {1, 2, 3, ...}$: $$ e^{-n} \sum_{k=0}^n \frac{n^k}{k!} \leq e^{-n} \sum_{k=0}^{\infty} \frac{n^k}{k!} = 1$$ and so the limit should be less than or equal to 1. Intuitively, we should expect it to be strictly less than 1 because we are not chopping the infinite sum at a point where it is close to $e^n$. Note that $n^k>k!$ when $n\geq k$. — Michael, Nov 07 '19 at 22:05

score 2 · Answer 1 · answered Nov 07 '19 at 22:04

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 $n \to \infty$.

score 2 · Answer 2 · answered Nov 07 '19 at 22:30

2

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 $1/2$.

answered Nov 07 '19 at 22:30

TBTD

3,516

Is CLT really necessary here? All you need is that the average of Pois$(1)$ random variables becomes very symmetric about $1$. – mathworker21 Nov 07 '19 at 22:33
Elaborate. That is a very vague statement. – TBTD Nov 07 '19 at 23:20
naw ${}{}{}{}{}$ – mathworker21 Nov 07 '19 at 23:27
@TBTD : Nice answer (I +1d). It was not obvious to me to use CLT. – Michael Nov 08 '19 at 00:03
This is the shortest solution. +1 – Paramanand Singh Nov 08 '19 at 03:21

grand_chat · Accepted Answer · 2019-11-08T01:08:55.077

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 $$ \lim_{n\to\infty}\left(1+\frac xn\right)^n=e^{x}\tag4 $$

edited Nov 08 '19 at 01:08

answered Nov 07 '19 at 22:50

grand_chat

38,951

The examples that you gave really help! Thank you so much! – Nov 08 '19 at 00:09
Sorry if this is too simple, but in one sentence, can we say my argument is wrong because the expansion of $e^x$ is only for real $x$? – Nov 08 '19 at 00:37
1

@BornaAhmadzade You could say the expansion is valid for any fixed $x$, where 'fixed' means not depending on $n$. – grand_chat Nov 08 '19 at 01:07
Very nice answer without the use of the formal term "bound variable". +1 – Paramanand Singh Nov 08 '19 at 03:19
@grand_chat I wouldn't say (2) is false, since it doesn't even make sense – mathworker21 Nov 08 '19 at 23:24

score 0 · Answer 4 · answered Nov 07 '19 at 22:19

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 $e^n / 2$.

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

4 Answers4