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

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geerky42
 one year ago
Best ResponseYou've already chosen the best response.2It is true for \(n=1\) and \(n=2\).

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0This site is terrible \[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(1)^kn^{ki}}{k!(ni)!}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes @Kainui i completely agree :P

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2I stumped this while solving big kid club lol @Kainui

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait no lol i read ur question in some other way :P sorry its not zeta

rational
 one year ago
Best ResponseYou've already chosen the best response.0seems to be true http://jsfiddle.net/ganeshie8/ykrpkweu/

dan815
 one year ago
Best ResponseYou've already chosen the best response.0is it enuff if i show it to be true for the integral analog of it

dan815
 one year ago
Best ResponseYou've already chosen the best response.0part of e^n i shud say in this case

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so its like uare adding higher and higher orders of e^n together

dan815
 one year ago
Best ResponseYou've already chosen the best response.0summation of 1/factorials have to with gamma function again

dan815
 one year ago
Best ResponseYou've already chosen the best response.0or u can think about it like adding different orders of e^1 now

dan815
 one year ago
Best ResponseYou've already chosen the best response.0higher and higher degeree approximations of e^1 being adding together

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2Not sure how approximation of \(e^n\) helps us?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where have u seen this formula @geerky42 ? or u just created it ?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2I kind of created it. I tried to solve @Kainui 's "the big kid's club" and I got stuck on closetolast step, then I saw solution, and bam! I got this formula.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2I 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...

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2Not sure how approximation of \(e^n\) helps us?
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