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