watchmath
  • watchmath
Interesting problem! Let f(n) be the number of zeros in n! Prove that lim f(n)/n = 1/4 (as n approaches infinity)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
We have the number of zeroes in n! :\[\sum_{i}^{}\lfloor n/5^{i} \rfloor\] this finds the "5's" and we have 5*2=10 which will give a zero. This is a start I geuss.
watchmath
  • watchmath
Great!
anonymous
  • anonymous
still trying to think about why the quotient goes to 1/4

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watchmath
  • watchmath
Now if we look at that series. Is there any relation between that "kind" of sereis(after we divide by n) and 1/4 ?
anonymous
  • anonymous
I think we can approximate it as the geometric series: \[\sum_{i=1}^{n}(1/5)^{i}\] which has sum 5/4-1=1/4 when n goes to infinity.
watchmath
  • watchmath
:) Now we just need to make it rigorous by figuring out how to open up that greatest integer function.
anonymous
  • anonymous
show that the error goes to zero when we approximate this way?
watchmath
  • watchmath
yes Note that the upper limit of the summation is \(\log_5n\), and this upper limit is somehow important.
anonymous
  • anonymous
Well in the approximation the upper limit is infinity, so the excess that is being added is: \[\sum_{i=(\log_{5} n)+1 }^{\infty}(1/5)^{i}\] this goes to 0 as n goes to infinity. Is this right? Also I'm trying to think of how to deal with the error from roundoff using the floor function.
watchmath
  • watchmath
Now quite because \(\lfloor n/5^i\rfloor/n\neq 1/5^i \) in general.

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