Question

Here's a fun question.\[S_n=\sum_{k=1}^n\left(\frac kn\right)^n\\\lim_{n\to\infty}S_n=\,\,?\]Note: This is a challenge. I have the answer/method if you want to see it.

Answer 1

you remind me of @FoolForMath :C

Answer 2

I was trying to fill in his absence. :c

Answer 3

it still feels different...his felt like you get humiliated because a "fool" is asking hard questions

in yours i feel skeptic...for obvious reasons

Answer 4

Ahaha.

Answer 5

I'd like to note that this question is "answerable" using techniques learned in basic calculus, but a more rigorous treatment can be provided using analysis.

Answer 6

1.58197

Answer 7

You can express the answer in terms of natural numbers and \(e\).

Answer 8

But, yes, you are correct. And apologies for multiple posting.

Answer 9

@KingGeorge You would be interest?

Answer 10

\[\frac{e}{e-1}\]

Answer 11

Equivalent into \[\sum_{k=0}^{\infty}e^{-k}\]

Answer 12

That's right. Do you have a proof written down, or would you like to see it?

Answer 13

I think the key is to notice that\[\lim_{n\to\infty}\left(\frac{n-m}{n}\right)^n=e^{-m}\]

Answer 14

Well, it doesn't matter. I'll post the proof: http://mathdl.maa.org/images/cms_upload/Holland-MMz-201039490.pdf

Answer 15

\[
\frac{e}{e-1.}\approx 1.5819
\]