- #1

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## Main Question or Discussion Point

Cute solution.

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

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

- Thread starter ZioX
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- #1

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Cute solution.

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

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

- #2

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- #3

StatusX

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- #4

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- #5

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Sum begins at n, which is going to infinity.

- #6

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so the limit becomes

lim n-> inf

1-e^{-n)*s

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

- #7

Gib Z

Homework Helper

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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 :)

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 :)

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- #8

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

I can give you guys a hint: I could've posted this in the probability forum.

- #9

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- #10

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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 \frac{x^n}{n!})[/tex]

[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]

- #11

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

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- #12

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Yes.

CLT for the win.

My statistician friend gave me the problem, and I hacked away at it for a good 5 hours and finally got a very non-trivial solution.

CLT for the win.

My statistician friend gave me the problem, and I hacked away at it for a good 5 hours and finally got a very non-trivial solution.

Last edited:

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