# Factorial Limit

1. May 22, 2015

### ThatDude

1. The problem statement, all variables and given/known data

So, I'm doing a series problem, and after applying the root test I end up with the following limit:

Lim as n--> inf of (n! / n^2) = infinity according to the solution manual.

I can't seem to understand this because the way I look at it, each term in the numerator is getting smaller [ (n)(n-1)(n-2)..] but each term in the denominator is staying constant as n^2. So shouldn't the limit equal zero?

2. Relevant equations

3. The attempt at a solution

2. May 22, 2015

### Ray Vickson

Google 'properties of factorial function' for example. You will see exactly what is going on for large n.

3. May 22, 2015

### pasmith

For $n \geq 3$ we have
$$\frac{n!}{n^2} \geq \frac{n(n-1)(n-2)}{n^2} = n - 3 + \frac{2}{n}.$$ The inequality follows from the fact that if $n = 3$ then $n! = n(n-1)(n-2)$ whilst if $n \geq 4$ then $(n-3) \times \dots \times 1 \geq 1$.

4. May 22, 2015

### Noctisdark

I recommand stirling approximation, it will be very appropriate for this case, it states that when n goes to infinity n! = √(2πn) * (n/e)^n, so try it !!

5. May 22, 2015

### vela

Staff Emeritus
You're looking at the wrong end in the numerator. Look at the n=3, n=4, and n=5 terms, for example. You have
\begin{align*}
\frac{3!}{3^2} &= \frac{3\cdot 2\cdot 1}{3\cdot 3} \\
\frac{4!}{4^2} &= \frac{4 \cdot 3\cdot 2\cdot 1}{4 \cdot 4} \\
\frac{5!}{5^2} &= \frac{5\cdot 4 \cdot 3\cdot 2\cdot 1}{5 \cdot 5}
\end{align*} Can you see the mistake in your reasoning?

Last edited: May 24, 2015
6. May 25, 2015

### ThatDude

So let's say I have the sequence n!(2^n) /(2n)!, how would I show that this converges to zero.

7. May 25, 2015

### Staff: Mentor

Start by expanding the (2n)! in the denominator to (2n)(2n -1) ...(n+1)(n!).