Prove that the limit as n->infinity of n^n/n! is infinity

[sphere1]

Homework Statement

Find the limit of (n^n)/n! as n approaches infinity

Relevant Equations

(n^n)/n!

I am self-studying Boas and this is a problem from Ch. 1.2. I have developed what I believe is an answer, but I'm not sure it's adequate. The general approach is to show that for all values of n > 1, n^n grows faster than n!, and therefore that (n^n)/n! approaches infinity as n approaches infinity:
1. n^n = n*n*..., and n! = n*(n-1)*(n-2)
2. n^n - n! = (n-n) + (n - (n-1)) + (n - (n-2)) + ... + (n-1) = 0 + 1 + ... + (n-1)
3. As n -> infinity, n^n - n! also approaches infinity, which implies that (n^n)/n! approaches infinity.

I realize that this problem could be solved by using L'Hôpital's rule, but I do not know how to differentiate a factorial. (I realize that it is possible, but I'm trying to solve this problem using my current toolkit.)

 

[Hill]

This line,

Homework Statement: Find the limit of (n^n)/n! as n approaches infinity
Relevant Equations: (n^n)/n!

2. n^n - n! = (n-n) + (n - (n-1)) + (n - (n-2)) + ... + (n-1) = 0 + 1 + ... + (n-1)

is wrong. Check for n=3. The left-hand side is 27-6=21. The right-hand side is 0+1+2=3.

 

[PeroK]

Your basic argument fails. For example:

As $n\to \infty$, we have $2n -n \to \infty$, yet $\frac{2n} n \to 2$.

 

[WWGD]

This may be an idea for a proof:
Say n=10.
10^10= 10(10)....(10) ; 10 times.
10!= 10(9)(8)...(5)...1.
10/5 =2.
______________
30^30=30(30)...(30) ; 30 times
30!= 30(29)(28)....(15)...1

30/15=2

You can do this doubling indefinitely.
Edit: I'm not giving all details because we're not supposed to do all work for students.

 

[fresh_42]

The convergence / divergence of sequences can often be proven by comparing a sequence with another one from which it is known whether it diverges or converges. You want to define divergence, so you need
$$
\displaystyle{\lim_{n \to \infty}\dfrac{n^n}{n!} \geq \lim_{n \to \infty} b_n}
$$
with a divergent series $(b_n)_{n\in \mathbb{N}}.$ What could we chose as $b_n$? (The simpler the better.)

 

[Orodruin]

The convergence / divergence of sequences can often be proven by comparing a sequence with another one from which it is known whether it diverges or converges. You want to define divergence, so you need
$$
\displaystyle{\lim_{n \to \infty}\dfrac{n^n}{n!} \geq \lim_{n \to \infty} b_n}
$$
with a divergent series $(b_n)_{n\in \mathbb{N}}.$ What could we chose as $b_n$? (The simpler the better.)

This is the best way if you can find a simple sequence $b_n$ (and it is not too hard to do so, a very simple example comes to mind). An alternative if one wants to brute force it is noting that $n^n/n! = n^{n-1}/(n-1)!$ and considering the logarithm of the sequence, i.e.,
$$
\ln(n^{n-1}/(n-1)!)
$$
and noting that this will diverge iff $n^n/n!$ diverges.

 

[nuuskur]

Homework Statement: Find the limit of (n^n)/n! as n approaches infinity
Relevant Equations: (n^n)/n!

I am self-studying Boas and this is a problem from Ch. 1.2. I have developed what I believe is an answer, but I'm not sure it's adequate. The general approach is to show that for all values of n > 1, n^n grows faster than n!, and therefore that (n^n)/n! approaches infinity as n approaches infinity:
1. n^n = n*n*..., and n! = n*(n-1)*(n-2)
2. n^n - n! = (n-n) + (n - (n-1)) + (n - (n-2)) + ... + (n-1) = 0 + 1 + ... + (n-1)
3. As n -> infinity, n^n - n! also approaches infinity, which implies that (n^n)/n! approaches infinity.

This is true, but how/why does it imply the limit of the ratio?

I realize that this problem could be solved by using L'Hôpital's rule, but I do not know how to differentiate a factorial. (I realize that it is possible, but I'm trying to solve this problem using my current toolkit.)

L'Hopital's rule applies for limits of differentiable functions. One would need to show $\frac{x^x}{x!} \to\infty$ as $x\to \infty$. If this is true, then the respective sequence also diverges. But there is a lot of theory in between that justifies this, so it would be preferable to avoid unnecessarily complicated approaches.

It suffices to note as you said
$$\frac{n^n}{n!} = \frac{n\cdot n\cdot \ldots\cdot n\cdot n}{n\cdot (n-1)\cdot \ldots 2\cdot 1 } = \frac{n\cdot n\ldots \cdot n}{n\cdot (n-1)\cdot \ldots \cdot 2} \cdot \frac{n}{1} \geqslant n,$$
because the left term is greater than $1$.

 

[PeroK]

Hint for $n > 2$: $n! = n(n-1) \dots (2)(1) < n^{n-1}$