Question

Prove:-
\[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k-1}{N}\right)^N = {1 \over e - 1}\]

Answer 1

I only know how to prove limits of sequences not series:(

Answer 2

what a cool series though..

Answer 3

yep!! this is very interesting!!

Answer 4

god we really dont know what we have on our hands with this math stuff.

Answer 5

I messed up on my first one...
\[\lim_{N\rightarrow\infty}\left(\frac{k-1}{N}\right)^N=\lim_{N\rightarrow\infty}\left(1+\frac{k-1-N}{N}\right)^N=e^{k-1-N}.\]

Answer 6

you missed the sum. the approach is +ve ... BRB in 3 hrs

Answer 7

And you can use a geometric series from there to get \(\displaystyle \frac{e^{-1}}{1-e^{-1}}\)

Answer 8

I know, it's not very rigorous. I've only learned as much analysis as I've taught myself, so I'm still not very good with proofs...

Answer 9

change of limits on the sum is the trick!!

Answer 10

sorry.. what do we call that ... k ... counter as in loop

Answer 11

Are you allowed to take the limit through the sum, as I have done?

Answer 12

For example, to go from:\[\large\lim_{x\rightarrow\infty}\sum_{k=0}^\infty f(x)\overset{?}{=}\sum_{k=0}^\infty \lim_{x\rightarrow\infty}f(x)\]

Answer 13

this is equivalent to
lim n->inf (1 - 1/n)^n + (1 - 2/n)^n + (1 - 3/n)^n + ....

Answer 14

let's put a new counter
m = N - k + 1 <-- this counts from opposite.

Answer 15

Yeah, that's what I did too. I didn't know it was called a counter, though.

Answer 16

i don't know what it is called either ... may be some incremental variable.

Answer 17

in programming ... in loop ... it's called counter... :D

Answer 18

lol

Answer 19

gotta go ... bye