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- experimentX

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

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- experimentX

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

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- zzr0ck3r

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

- zzr0ck3r

what a cool series though..

- experimentX

yep!! this is very interesting!!

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- zzr0ck3r

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

- anonymous

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

- experimentX

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

- anonymous

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

- anonymous

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...

- experimentX

change of limits on the sum is the trick!!

- experimentX

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

- anonymous

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

- anonymous

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

- experimentX

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

- experimentX

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

- anonymous

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

- experimentX

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

- experimentX

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

- anonymous

lol

- experimentX

gotta go ... bye

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