Recent content by smb26

I guess I found the solution fact is ##X! < X^x## so we say that by Direct Comparison Test $$\frac{ln(n!)}{n^3 *ln(n)} < \frac {ln(n^n)}{n^3 *ln(n)} = \frac {n*ln(n)}{n^3 *ln(n)} = \frac {1}{n^2 } $$ since $$\frac {1}{n^2 } $$ is a convergent series of p=2, the $$Ʃ\frac{ln(n!)}{n^3...

The separation of ln(n!) will help me how ?

[ln((n+1)!)]/[ln(n+1)*(n+1)^3] * [ln(n)*n^3]/[ln(n!)] now at infinity n^3/(n+1)^3 will give 1 & ln(n)/ln(n+1) will give 1 I guess but what about ln((n+1)!)/ln(n!) what is this going to give & how to prove... Can this be done by Direct Comparison Test or Limit Comparison Test??

ln(n) is in the deonminator

1. Homework Statement ∞ [SIZE="6"]Ʃ ln(n!)/(n^3)*ln(n) n=2 2. Homework Equations Ratio test, which states: Lim as n tends to infinity of |(An+1)/An|=ρ would lead to convergent series if the limit is less than 1 it diverges if ρ is greater than 1 test is inconclusive if ρ=1...