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experimentX
Group Title
Prove:
\[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k1}{N}\right)^N = {1 \over e  1}\]
 2 years ago
 2 years ago
experimentX Group Title
Prove: \[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k1}{N}\right)^N = {1 \over e  1}\]
 2 years ago
 2 years ago

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zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
I only know how to prove limits of sequences not series:(
 2 years ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
what a cool series though..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yep!! this is very interesting!!
 2 years ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
god we really dont know what we have on our hands with this math stuff.
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
I messed up on my first one... \[\lim_{N\rightarrow\infty}\left(\frac{k1}{N}\right)^N=\lim_{N\rightarrow\infty}\left(1+\frac{k1N}{N}\right)^N=e^{k1N}.\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
you missed the sum. the approach is +ve ... BRB in 3 hrs
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
And you can use a geometric series from there to get \(\displaystyle \frac{e^{1}}{1e^{1}}\)
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.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...
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
change of limits on the sum is the trick!!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
sorry.. what do we call that ... k ... counter as in loop
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
Are you allowed to take the limit through the sum, as I have done?
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
For example, to go from:\[\large\lim_{x\rightarrow\infty}\sum_{k=0}^\infty f(x)\overset{?}{=}\sum_{k=0}^\infty \lim_{x\rightarrow\infty}f(x)\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this is equivalent to lim n>inf (1  1/n)^n + (1  2/n)^n + (1  3/n)^n + ....
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let's put a new counter m = N  k + 1 < this counts from opposite.
 2 years ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.1
Yeah, that's what I did too. I didn't know it was called a counter, though.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i don't know what it is called either ... may be some incremental variable.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
in programming ... in loop ... it's called counter... :D
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
gotta go ... bye
 2 years ago
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