experimentX
  • experimentX
Prove:- \[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k-1}{N}\right)^N = {1 \over e - 1}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zzr0ck3r
  • zzr0ck3r
I only know how to prove limits of sequences not series:(
zzr0ck3r
  • zzr0ck3r
what a cool series though..
experimentX
  • experimentX
yep!! this is very interesting!!

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zzr0ck3r
  • zzr0ck3r
god we really dont know what we have on our hands with this math stuff.
anonymous
  • 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
  • experimentX
you missed the sum. the approach is +ve ... BRB in 3 hrs
anonymous
  • anonymous
And you can use a geometric series from there to get \(\displaystyle \frac{e^{-1}}{1-e^{-1}}\)
anonymous
  • 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
  • experimentX
change of limits on the sum is the trick!!
experimentX
  • experimentX
sorry.. what do we call that ... k ... counter as in loop
anonymous
  • anonymous
Are you allowed to take the limit through the sum, as I have done?
anonymous
  • 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
  • experimentX
this is equivalent to lim n->inf (1 - 1/n)^n + (1 - 2/n)^n + (1 - 3/n)^n + ....
experimentX
  • experimentX
let's put a new counter m = N - k + 1 <-- this counts from opposite.
anonymous
  • anonymous
Yeah, that's what I did too. I didn't know it was called a counter, though.
experimentX
  • experimentX
i don't know what it is called either ... may be some incremental variable.
experimentX
  • experimentX
in programming ... in loop ... it's called counter... :D
anonymous
  • anonymous
lol
experimentX
  • experimentX
gotta go ... bye

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