Question

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

Answer 1

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.

Answer 2

Great!

Answer 3

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

Answer 4

Now if we look at that series. Is there any relation between that "kind" of sereis(after we divide by n) and 1/4 ?

Answer 5

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.

Answer 6

:)
Now we just need to make it rigorous by figuring out how to open up that greatest integer function.

Answer 7

show that the error goes to zero when we approximate this way?

Answer 8

yes
Note that the upper limit of the summation is \(\log_5n\), and this upper limit is somehow important.

Answer 9

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.

Answer 10

Now quite because \(\lfloor n/5^i\rfloor/n\neq 1/5^i \) in general.