## experimentX Group Title Prove:- $\lim_{N\rightarrow\infty}\sum_{k=1}^N\left(\frac{k-1}{N}\right)^N = {1 \over e - 1}$ one year ago one year ago

1. zzr0ck3r Group Title

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

2. zzr0ck3r Group Title

what a cool series though..

3. experimentX Group Title

yep!! this is very interesting!!

4. zzr0ck3r Group Title

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

5. Herp_Derp Group Title

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 Group Title

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

7. Herp_Derp Group Title

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

8. Herp_Derp Group Title

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 Group Title

change of limits on the sum is the trick!!

10. experimentX Group Title

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

11. Herp_Derp Group Title

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

12. Herp_Derp Group Title

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 Group Title

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

14. experimentX Group Title

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

15. Herp_Derp Group Title

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

16. experimentX Group Title

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

17. experimentX Group Title

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

18. Herp_Derp Group Title

lol

19. experimentX Group Title

gotta go ... bye