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

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

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.1what a cool series though..

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1yep!! this is very interesting!!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I messed up on my first one... \[\lim_{N\rightarrow\infty}\left(\frac{k1}{N}\right)^N=\lim_{N\rightarrow\infty}\left(1+\frac{k1N}{N}\right)^N=e^{k1N}.\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And you can use a geometric series from there to get \(\displaystyle \frac{e^{1}}{1e^{1}}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.1change of limits on the sum is the trick!!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are you allowed to take the limit through the sum, as I have done?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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
 4 years ago
Best ResponseYou've already chosen the best response.1this is equivalent to lim n>inf (1  1/n)^n + (1  2/n)^n + (1  3/n)^n + ....

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's what I did too. I didn't know it was called a counter, though.

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

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