geerky42
  • geerky42
How can I verify if this is true for \(n\in\mathbb N\)? \[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}\]
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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geerky42
  • geerky42
It is true for \(n=1\) and \(n=2\).
Kainui
  • Kainui
This site is terrible \[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}\]
anonymous
  • anonymous
yes @Kainui i completely agree :P

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More answers

geerky42
  • geerky42
I stumped this while solving big kid club lol @Kainui
anonymous
  • anonymous
wait no lol i read ur question in some other way :P sorry its not zeta
rational
  • rational
seems to be true http://jsfiddle.net/ganeshie8/ykrpkweu/
dan815
  • dan815
is it enuff if i show it to be true for the integral analog of it
dan815
  • dan815
|dw:1441227726264:dw|
dan815
  • dan815
|dw:1441227959575:dw|
dan815
  • dan815
|dw:1441228057588:dw|
dan815
  • dan815
part of e^-n i shud say in this case
dan815
  • dan815
so its like uare adding higher and higher orders of e^-n together
dan815
  • dan815
summation of 1/factorials have to with gamma function again
dan815
  • dan815
or u can think about it like adding different orders of e^1 now
dan815
  • dan815
higher and higher degeree approximations of e^1 being adding together
dan815
  • dan815
g2g mow the lawn
anonymous
  • anonymous
:) @dan815
geerky42
  • geerky42
Not sure how approximation of \(e^n\) helps us?
anonymous
  • anonymous
where have u seen this formula @geerky42 ? or u just created it ?
geerky42
  • 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.
geerky42
  • 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...
anonymous
  • anonymous
:O thats cool
geerky42
  • geerky42
@dan815 ?
geerky42
  • geerky42
Not sure how approximation of \(e^n\) helps us?

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