experimentX 2 years ago Prove:- $\lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k-1}{N}\right)^N = {1 \over e - 1}$

1. zzr0ck3r

I only know how to prove limits of sequences not series:(

2. zzr0ck3r

what a cool series though..

3. experimentX

yep!! this is very interesting!!

4. zzr0ck3r

god we really dont know what we have on our hands with this math stuff.

5. Herp_Derp

I messed up on my first one... $\lim_{N\rightarrow\infty}\left(\frac{k-1}{N}\right)^N=\lim_{N\rightarrow\infty}\left(1+\frac{k-1-N}{N}\right)^N=e^{k-1-N}.$

6. experimentX

you missed the sum. the approach is +ve ... BRB in 3 hrs

7. Herp_Derp

And you can use a geometric series from there to get $$\displaystyle \frac{e^{-1}}{1-e^{-1}}$$

8. Herp_Derp

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...

9. experimentX

change of limits on the sum is the trick!!

10. experimentX

sorry.. what do we call that ... k ... counter as in loop

11. Herp_Derp

Are you allowed to take the limit through the sum, as I have done?

12. Herp_Derp

For example, to go from:$\large\lim_{x\rightarrow\infty}\sum_{k=0}^\infty f(x)\overset{?}{=}\sum_{k=0}^\infty \lim_{x\rightarrow\infty}f(x)$

13. experimentX

this is equivalent to lim n->inf (1 - 1/n)^n + (1 - 2/n)^n + (1 - 3/n)^n + ....

14. experimentX

let's put a new counter m = N - k + 1 <-- this counts from opposite.

15. Herp_Derp

Yeah, that's what I did too. I didn't know it was called a counter, though.

16. experimentX

i don't know what it is called either ... may be some incremental variable.

17. experimentX

in programming ... in loop ... it's called counter... :D

18. Herp_Derp

lol

19. experimentX

gotta go ... bye