Divergent Sequence Homework: Determine Convergence

[titasB]

Homework Statement

Determine whether the sequence is convergent or divergent

Homework Equations

a_n = {nⁿ / n! }

The Attempt at a Solution

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

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

⇒ as n → ∞ , a_n → ∞

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 :)

 

[jbunniii]

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

 

[titasB]

So something like this:

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

a_n =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,

∞ ≥ a_n ≥ n

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

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

 

[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?

 

[titasB]

nⁿ/ n! ≥ n

 

[jbunniii]

nⁿ/ n! ≥ n

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##.

 

[titasB]

Awesome thanks so much!