## watchmath 5 years ago Interesting problem! Let f(n) be the number of zeros in n! Prove that lim f(n)/n = 1/4 (as n approaches infinity)

1. 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.

2. watchmath

Great!

3. anonymous

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

4. 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 ?

5. 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.

6. watchmath

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

7. anonymous

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

8. watchmath

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

9. 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.

10. watchmath

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