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

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.

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!

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!

@_@ - 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... -_-.

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

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.)