# Divergent Sequence

1. Apr 23, 2015

### titasB

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

Determine whether the sequence is convergent or divergent

2. Relevant equations

an = {nn / n! }

3. The attempt at a solution

an = ( n . n. n ... n ) / 1 . 2 . 3 ... n )

⇒ an =n [ ( n . n. n ... n ) / 1 . 2 . 3 ... n ) ]

⇒ as n → ∞ , an → ∞

This is further confirmed by finding the values for the first 6 terms : 1, 2, 4.5, 10.67,26.04,64.8...

I'm not sure if this is sufficient however. Can we explain this using limits? If so wouldn't that give an indeterminant form? Of course the factorial on the denominator will always be smaller than nn but can this be shown using limits? Any help would be nice :)

2. Apr 23, 2015

### jbunniii

Hint: Can you show that
$$\frac{n^n}{n!} \geq n$$

3. Apr 24, 2015

### titasB

So something like this:

an = ( n . n. n ... n ) / 1 . 2 . 3 ... n )

an =n [ ( n . n. n ... n ) / 1 . 2 . 3 ... n ) ]

we can see the expression in the brackets approach infinity since the numerator is much larger than the denominator so,

∞ ≥ an ≥ n

and since n → ∞ this gives : ∞ ≥ an ≥ ∞

and so, an → ∞ as n → ∞ by squeeze theorem ⇒ the sequence is divergent

4. Apr 24, 2015

### jbunniii

I think you need to be more precise. How many $n$'s are there in the numerator of your second expression? Can you quantify "the numerator is much larger than the denominator"?

Consider factoring as follows:
$$\frac{n^n}{n!} = \left(\frac{n}{1}\right) \left[\left(\frac{n}{2}\right) \cdots \left(\frac{n}{n-1}\right)\left(\frac{n}{n}\right)\right]$$
Now look at the factors enclosed in the square brackets. Each factor is $\geq 1$, so the product in the square brackets is $\geq 1$. What can you conclude?

5. Apr 24, 2015

### titasB

nn/ n! ≥ n

6. Apr 24, 2015

### jbunniii

Right, now the rest of the argument works as you stated. The inequality shows that as $n \to \infty$, the larger expression $n^n / n!$ must also diverge to $\infty$.

7. Apr 24, 2015

### titasB

Awesome thanks so much!