Has anyone seen this problem before?

Why don't you answer the (my) question in the math Q&A topic in the general math section, then you could post this as a problem.

 

that is the formula for e approximation. i did that in programming, but not approaching infinity, it should be real number.. i might be wrong.

 

the series is just e^n- s, where s is the partial sum of formal power series e^n truncated when i=n

so the limit becomes
lim n-> inf
1-e^{-n)*s

s goes to e^{n} and the whole thing goes to zero...?

 

Yea i get the same answer as tim, The whole thing goes to zero.

Edit: Working-

[tex]\lim_{n\to\infty}e^{-n}\sum_{i=n}^{\infty}\frac{n^i}{i!}[/tex]

[tex]\lim_{n\to\infty}e^{-n}(e^n - \sum_{i=0}^n \frac{x^n}{n!})[/tex]

[tex]\lim_{n\to\infty}(1 - \frac{\sum_{i=0}^n \frac{x^n}{n!}}{e^n})[/tex]

[tex]1 - \frac{\lim_{n\to\infty}\sum_{i=0}^n \frac{x^n}{n!}}{\lim_{n\to\infty} e^n}[/tex]

[tex]1 - \frac{\sum_{n=0}^{\infty} \frac{x^n}{n!}}{e^n}[/tex]

[tex]1 - \frac{e^n}{e^n}[/tex]

Which is zero. I think i included enough obvious steps for everyone to follow :)

 

if you graph this, this equation the domain is approximately (-inf, inf), while its range is the same which is infinity. ziox is right it's not zero, but if you take n as real number the equation solves the e approximation in e^n.

 

First correction
[tex]\lim_{n\to\infty}e^{-n}\sum_{i=n}^{\infty}\frac{n^i}{i!}[/tex]

[tex]\lim_{n\to\infty}e^{-n}(e^n - \sum_{i=0}^{n-1} \frac{n^i}{i!})[/tex]

 

By the way, is the limit = 0.5 = P[Z>0], Z~N(0,1).