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From: Thomas on 13 Jun 2010 02:26
It does diverge because of the fact that 1/(x^((x+1)/x)) approaches 1/x as x approaches infinity, the Limit Convergence Theorem and the fact that the sum from x=1 to infinity of 1/x diverges.
From: Ray Vickson on 13 Jun 2010 12:53
On Jun 13, 3:26 am, Thomas <thomasinthail...(a)gmail.com> wrote:
> It does diverge because of the fact that 1/(x^((x+1)/x)) approaches 1/x as x approaches infinity, the Limit Convergence Theorem and the fact that the sum from x=1 to infinity of 1/x diverges.

Let f(n) = 1/[n^((n+1)/n))]. What is limit_{n --> infinity} n*f(n)?
Remembering the _definition_ of "limit", what does the above result
tell you?

R.G. Vickson
From: Ray Vickson on 13 Jun 2010 20:20
On Jun 13, 9:53 am, Ray Vickson <RGVick...(a)shaw.ca> wrote:
> On Jun 13, 3:26 am, Thomas <thomasinthail...(a)gmail.com> wrote:
>
> > It does diverge because of the fact that 1/(x^((x+1)/x)) approaches 1/x as x approaches infinity, the Limit Convergence Theorem and the fact that the sum from x=1 to infinity of 1/x diverges.
>
> Let f(n) = 1/[n^((n+1)/n))]. What is limit_{n --> infinity} n*f(n)?
> Remembering the _definition_ of "limit", what does the above result
> tell you?
>
> R.G. Vickson

To clarify: your stated notion that f(n) "approaches" 1/n is vague and
not very useful, at least in some interpretations. However, the notion
that n*f(n) has some computable limit is very useful indeed.

R.G. Vickson
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