You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

Recently someone came up with a question on a calculus yahoo group that no one replied:

Considering the sequence

`a(n)=(n!)^(1/n) ; n∈N`

Prove that

```
lim a(n) = +oo
n->+oo
```

..any suggestions?

Offline

**ajp3****Member**- Registered: 2005-08-25
- Posts: 9

First, recall the theorem ( a(n) -> 0 ) => ( (a(n))^(1/n) -> 0 ) , where n -> ∞ in both limits. (the proof of this is pretty straightforward, but I can do it if you like)

Then note the well known fact that (x^n)/(n!) -> 0 as n -> ∞; put this together with the above theorem and get:

x/((n!)^(1/n)) -> 0 as n -> ∞.

Since x is constant in the limit, we must have:

(n!)^(1/n) -> ∞ as n -> ∞, and it's positive infinity since n! > 1 for all n in N.

[replaced ? with ∞ for you - mathsisfun]

*Last edited by MathsIsFun (2005-08-26 10:34:48)*

Offline

**ajp3****Member**- Registered: 2005-08-25
- Posts: 9

darn... I was trying to be too fancy.... replace all the question marks in that above post with the symbol 'infinity' (or '+infinity', if you prefer)

Offline

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

ok thanks!

Offline

Pages: **1**