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#1 2005-08-25 10:25:37

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

limits - try this one!

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?

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#2 2005-08-25 13:15:52

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

Re: limits - try this one!

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)

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#3 2005-08-25 13:17:05

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

Re: limits - try this one!

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)

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#4 2005-08-26 06:22:38

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

Re: limits - try this one!

ok thanks!

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