## geerky42 one year ago How can I verify if this is true for $$n\in\mathbb N$$? $\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}$

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1. geerky42

It is true for $$n=1$$ and $$n=2$$.

2. Kainui

This site is terrible $\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}$

3. anonymous

yes @Kainui i completely agree :P

4. geerky42

I stumped this while solving big kid club lol @Kainui

5. anonymous

wait no lol i read ur question in some other way :P sorry its not zeta

6. rational

seems to be true http://jsfiddle.net/ganeshie8/ykrpkweu/

7. dan815

is it enuff if i show it to be true for the integral analog of it

8. dan815

|dw:1441227726264:dw|

9. dan815

|dw:1441227959575:dw|

10. dan815

|dw:1441228057588:dw|

11. dan815

part of e^-n i shud say in this case

12. dan815

so its like uare adding higher and higher orders of e^-n together

13. dan815

summation of 1/factorials have to with gamma function again

14. dan815

or u can think about it like adding different orders of e^1 now

15. dan815

higher and higher degeree approximations of e^1 being adding together

16. dan815

g2g mow the lawn

17. anonymous

:) @dan815

18. geerky42

Not sure how approximation of $$e^n$$ helps us?

19. anonymous

where have u seen this formula @geerky42 ? or u just created it ?

20. geerky42

I kind of created it. I tried to solve @Kainui 's "the big kid's club" and I got stuck on close-to-last step, then I saw solution, and bam! I got this formula.

21. geerky42

I applied Binomial Theorem and took tedious path, so I sincerely hope this formula is true, because my solution is like 2 whole pages long and otherwise it'd be hard to hunt for mistake lol...

22. anonymous

:O thats cool

23. geerky42

@dan815 ?

24. geerky42

Not sure how approximation of $$e^n$$ helps us?