Here's the question you clicked on:
experimentX
Prove:- \[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k-1}{N}\right)^N = {1 \over e - 1}\]
I only know how to prove limits of sequences not series:(
what a cool series though..
yep!! this is very interesting!!
god we really dont know what we have on our hands with this math stuff.
I messed up on my first one... \[\lim_{N\rightarrow\infty}\left(\frac{k-1}{N}\right)^N=\lim_{N\rightarrow\infty}\left(1+\frac{k-1-N}{N}\right)^N=e^{k-1-N}.\]
you missed the sum. the approach is +ve ... BRB in 3 hrs
And you can use a geometric series from there to get \(\displaystyle \frac{e^{-1}}{1-e^{-1}}\)
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...
change of limits on the sum is the trick!!
sorry.. what do we call that ... k ... counter as in loop
Are you allowed to take the limit through the sum, as I have done?
For example, to go from:\[\large\lim_{x\rightarrow\infty}\sum_{k=0}^\infty f(x)\overset{?}{=}\sum_{k=0}^\infty \lim_{x\rightarrow\infty}f(x)\]
this is equivalent to lim n->inf (1 - 1/n)^n + (1 - 2/n)^n + (1 - 3/n)^n + ....
let's put a new counter m = N - k + 1 <-- this counts from opposite.
Yeah, that's what I did too. I didn't know it was called a counter, though.
i don't know what it is called either ... may be some incremental variable.
in programming ... in loop ... it's called counter... :D