Find the limit (fairly difficult, may be related to Maclaurin series of e^x)?

Find the given limit:

lim { e^-n * (1 / 0! + n / 1! + n^2 / 2! + n^3 / 3! + ... + n^n / n!) }
as n goes to infinity.

This was a problem on a friend's exam a while back. I ended up evaluating the limit after he and I talked about it, but I had to use some decently high-level tools on the way. If anyone is interested, I'll give the sketch of what I did, but I'm curious to see if anyone can solve it differently.

2012-08-28T21:34:27Z

Both my friend and I thought it was 1 at first. The problem is that the series for e^x doesn't converge uniformly, so you can't just use the convergence of the Maclaurin series. The limit, for those who are still interested, is 1/2 (I debated putting it in the question to begin with...).

It may very well be unsolvable using just ordinary calculus tools. I would still welcome an elementary-ish solution, but mostly I'm just wondering if there's a different proof from my own.

2012-08-28T21:51:28Z

alwbsok: You forgot the (n+1)st derivative term in Taylor's theorem. If you were to fix this, then the resulting limit you would want to show (for the error bound) is
n^(n+1) / (n+1)! - 0,
which isn't true anymore.

Thanks for the serious answer (it's odd how rare they are on here sometimes!).

Waiting to see if there's another proof!

2012-08-28T21:54:02Z

Of course, I meant
"n^(n+1) / (n+1)! - 0"
with the arrow for the limit... hurrah for the inability to correct typos in questions!

2012-08-28T22:01:04Z

@_@ - Okay, now Y!A is just messing with me. I chalked it up to my being tired that I missed the arrow the first time, but the second I know I put it there... -_-.

2012-08-29T17:21:46Z

gianlino: I do agree your limit is equivalent, but I wasn't able to find how to show it...

And since you asked, my proof goes something like this:
If X_1, ... , X_n are independent Poisson random variables with mean 1, then the sum
S_n = X_1 + ... + X_n
is a Poisson random variable with mean n.

Then the limit we want is the limit of
P( S_n

2012-08-29T17:25:28Z

That's bizarre... it keeps cutting out random parts of my details...

The limit is the limit of
P( S_n less than or equal to n ),
which by the central limit theorem is
P( Z less than or equal to 0 ) = 1/2
(where Z is a std normal r.v.)

gianlino2012-08-29T07:56:33Z

Favorite Answer

It is equivalent to show that

lim { e^-n * (sum [k<= n] n^k/k! - sum [k > n] n^k/k! ) } = 0.

If you cut at n/2, 3n/2 and group symmetrically the terms in between, each part goes to 0 and it is not hard to show.

Taylor is irrelevant here. After you use the power series expansion of e^n, you just perform elementary estimates.

Btw what is, in few words, your own proof?

edit: Of course using the CLT theorem for that seems a little too much but why not? Did you try the cutting and grouping I suggested?

alwbsok2012-08-28T20:55:54Z

Right, I know what I screwed up. I should have checked my first step. I misapplied Taylor's theorem. The inequality I needed was:

|e^x - (1 / 0! + x / 1! + x^2 / 2! + ... + x^n / n!)| <= e^x * x^(n + 1) / (n + 1)!

which would spoil the convergence. Let me think about this.

Ah, I see you caught that too.

Anonymous2012-08-28T20:24:56Z

The series is the expansion of e^n, so the limit is 1 (e^0)

Popo2012-09-01T22:27:59Z

As always, math stack exchange has several *excellent* proofs. (Most of them are methodically quite similar, though.)

http://math.stackexchange.com/questions/160248/compute-the-limit-lim-n-rightarrow-infty-e-n-sum-limits-k-0n-frac

I was disturbed by the possibility that this could be a high-school calculus exercise...