Convergence of a series: sigma, n=2 to infty, 1/(ln n)^ln n

[grapz]

Test the series for convergence or divergence.

Sigma from n = 2 --> infinity of 1 / ( ln n ) ^ ln n

I think we have to use comparision test, but i am not sure what function to compare it to

 

[royhaas]

Re: Convergence of a series

You can use the fact that \(\displaystyle 2 \le ln(n) \le n , n \ge 8\).

 

[Dr. Flim-Flam]

Converges by Integral Test.

Note: for x >=2, 1/[ln(x)]^ln(x) is a decreasing, positive, and continuous function (its derivative is always negative) and

integral 1/[ln(x)]^ln(x), x,2,Inf = 5.03717 (TI=89), hence Sum a_n,2,inf converges.