G grapz Junior Member Joined Jan 13, 2007 Messages 80 Apr 10, 2008 #1 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
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
R royhaas Full Member Joined Dec 14, 2005 Messages 832 Apr 10, 2008 #2 Re: Convergence of a series You can use the fact that \(\displaystyle 2 \le ln(n) \le n , n \ge 8\).
D Dr. Flim-Flam Junior Member Joined Oct 10, 2007 Messages 108 Apr 12, 2008 #3 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.
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.
