Interesting problem! Let f(n) be the number of zeros in n! Prove that lim f(n)/n = 1/4 (as n approaches infinity)

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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
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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.
Great!
still trying to think about why the quotient goes to 1/4

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Now if we look at that series. Is there any relation between that "kind" of sereis(after we divide by n) and 1/4 ?
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.
:) Now we just need to make it rigorous by figuring out how to open up that greatest integer function.
show that the error goes to zero when we approximate this way?
yes Note that the upper limit of the summation is \(\log_5n\), and this upper limit is somehow important.
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.
Now quite because \(\lfloor n/5^i\rfloor/n\neq 1/5^i \) in general.

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