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Prove:
\[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k1}{N}\right)^N = {1 \over e  1}\]
 one year ago
 one year ago
Prove: \[ \lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k1}{N}\right)^N = {1 \over e  1}\]
 one year ago
 one year ago

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zzr0ck3rBest ResponseYou've already chosen the best response.1
I only know how to prove limits of sequences not series:(
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
what a cool series though..
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yep!! this is very interesting!!
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
god we really dont know what we have on our hands with this math stuff.
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
you missed the sum. the approach is +ve ... BRB in 3 hrs
 one year ago

Herp_DerpBest ResponseYou've already chosen the best response.1
And you can use a geometric series from there to get \(\displaystyle \frac{e^{1}}{1e^{1}}\)
 one year ago

Herp_DerpBest 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...
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
change of limits on the sum is the trick!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sorry.. what do we call that ... k ... counter as in loop
 one year ago

Herp_DerpBest ResponseYou've already chosen the best response.1
Are you allowed to take the limit through the sum, as I have done?
 one year ago

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

experimentXBest 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 + ....
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
let's put a new counter m = N  k + 1 < this counts from opposite.
 one year ago

Herp_DerpBest 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i don't know what it is called either ... may be some incremental variable.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
in programming ... in loop ... it's called counter... :D
 one year ago
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