How do you prove that ?

Oh man, you totally spoiled it! This isn’t how you ask the question, come on!

You ask it like this: What is

\displaystyle \lim_{n\to\infty} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} \,\,\text{ ?}

And then you sit back and enjoy the spectacle of everyone saying that it’s obviously 1, and they explain why, and it’s so clear that it doesn’t even require proof but you ask them to prove it anyway, and they try, and they fail (or worse: succeed), and you ask them if they still think it’s 1, and they say yeah but some of the confidence is gone, and you play them for just as long as it pleases you until you inform them they are exactly 100% off.

So, why does everyone think this limit equals 1, and why is that not true, and why is it actually \frac{1}{2}?

Well, for n very large, the sum \sum_{k=0}^n \frac{n^k}{k!} is about e^n. Right? It’s just the Taylor series of the exponential function. So then we multiply that by e^{-n} and get just about 1, and as we let n grow this just becomes more and more accurate, so the limit has to be simply 1. I mean it has to.

Right?

Wrong.

So what’s wrong here? Well, you may feel a little uncomfortable about that Taylor business. I mean, sure, the sum

\displaystyle \sum_{k=0}^n \frac{x^k}{k!}

Is a partial sum of the Taylor series of e^x, so its limit for fixed x as n\to\infty is indeed e^x. But here we are doing something a little suspicious: we’re asking n to perform double duty as both the summation range and the variable of the power series.

One thing, at any rate, should be clear: the statement

\displaystyle e^n=?=\lim_{n\to\infty}\sum_{k=0}^{n}\frac{n^k}{k!}

makes no sense at all. The variable n is free on the left side and bound on the right.

Ok. So that naive interpretation is out the window. How do we evaluate that limit?

There’s a beautiful, masterful way to solve this that’s almost sleight of hand. It goes like this: this is precisely the limiting probability of a random Poisson variable with parameter \lambda=n to be less than its expectation. Such a variable is distributed just like the sum of n independent Poisson variables with parameter \lambda=1, and such a sum (normalized by its variance, \sqrt{n}, which doesn’t matter here) converges in distribution to a normal distribution. What’s the probability that a normal random variable be less than its mean? Why it’s \frac{1}{2}, of course. Done. QED.

Wait, what?

Yes, really. If you know about the Central Limit Theorem, this is exactly what it says if you take random variables X_1,X_2,\ldots each of which is Poisson(1). Good, routine exercise in applying CLT to random Poisson variables.

But what if you don’t know about the CLT or it just didn’t occur to you to interpret this limit as a probability?

Then this is, honestly, quite a hard problem. The CLT is a powerful theorem which hides quite a bit of theory, offering it virtually for free. Without it, you’re on your own here, and I don’t know a really easy way to prove this. Some clever integral manipulations and transformations are needed.