I am struggling with this one. For what values for p does the series converge?
$\displaystyle \sum_{n=1}^\infty $$\displaystyle ln(n)/n^p$
I have tried the ratio test but I keep getting 1(inconclusive)
I am not sure by what you typed but it looks like you said its true for any p > 0 but when p=1 then $\displaystyle ln(n)/n$ and that is divergent. But when p = 2 or anything greater it looks like it is convergent. I need a way to show this. I think that if I can show that when p =2 that that is convergent than by comparison any p > 2 is convergent but that may be incorrect as well???
Don't use your TI-89! Use your brain! Distribute $\displaystyle n^{-p}$ and notice what happens.
Here is one final way you could prove it, since your series is decreasing and positive after some value of N (use the inequality Ifirst gaave you to prove that) then your series covnerges iff $\displaystyle \sum_{n=1}^{\infty}2^n\cdot\frac{\ln\left(2^n\righ t)}{2^{np}}=\sum_{n=1}^{\infty}\frac{\ln(2)n}{2^{( p-1)n}}$
Now think what values of p does this serise converges...this time its pretty apparent.