Fact of Factorials

Number Theory Level 4

Let $f(n) : \mathbb{Z}^+ \to \mathbb{Z}$ be a function which gives the number of trailing zeros in $n!$ .

Then find,

$\lim_{n\to \infty} \frac{f(n)}{n}$

Details and Assumptions

For example, if $n=3$ , then $3! = 6$ , so $f(3) = 0$ ; if $n=7$ , then $7! = 5040$ , so $f(7) = 1$

Image Credit: Wikimedia Factorial Interpolation

The answer is 0.25.

1 solution

Rajath Naik
Mar 5, 2015

Here the function f(n) can be expanded as n/5 + n/5² +n/5³..........infinity..from the definition of trailing numbers. So when this summation is divided by n, the entire function is independent of n. And this function now becomes a GP summation expansion till infinity,with first term a=1/5 and common ratio r=1/5. .hence the infinite sum is 0.25

Note that the terms should be in the floor function, so it is not exactly equal to $\frac{n}{4}$ . You will need to account for this error term, which is at most $4 \log_5 n$ and hence really small compared to $\frac{n}{4}$ .

Calvin Lin Staff - 6 years, 3 months ago

How did you get that error ?

Vishal Yadav - 4 years, 2 months ago

Write it out.

$\sum_k \frac{ n } { 5^k } - \lfloor \frac{ n } { 5 ^ k } \rfloor \leq \ldots$

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – You mean $\sum_k\frac{n}{5^k} - \floor{\frac{n}{5^k}} \leq \sum_k\frac{n+1}{5^k} - \floor{\frac{n+1}{5^k}}\leq$ ...

What do I do with that ?

Vishal Yadav - 4 years, 2 months ago

@Vishal Yadav – Bound each term directly.

E.g. each term is less than 1, and there are at most $\log_5 n$ terms. Hence the error is at most $\log _ 5 n$ .

All we need is a crude bound.

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – Thanks. But I'm haven't learned enough to understand what it means...Will give a try after sometime..

Vishal Yadav - 4 years, 2 months ago

Fact of Factorials

Image Credit: Wikimedia Factorial Interpolation

The answer is 0.25.

1 solution

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