for what values of p does.... 1. The problem statement, all variables and given/known data Find the values of p for which the series is convergent: 2. Relevant equations There are two separate problems I'm having problems with here. The first is: n=1 to n=∞ ∑ 1/(n(lnn)^p) The second is: n=1 to n=∞ ∑ ln(n)/(n^p) 3. The attempt at a solution Okay, I'm almost certain both of them converge for all values p>1. I'm having trouble proving it though. Question 1: On this one, I first took the sister function f(x)=1/x(lnx)^p and integrated it using u-substitution: ∫ (1/x(lnx)^p )dx u=lnx du=(1/x)dx ∫u^(-p)du = (u^(-p+1)) / (-p+1) which I rewrote as: (u^(1-p)) / (1-p) then, I took the limit lim n→∞ (u^(1-p)) / (1-p), looking for where it would converge. I didn't really do any editing here, I just assumed p had to be greater than 1 so that the bottom was not negative, and not equal to 0, which would be undefined. Likewise, the top would have a negative exponent which would make the ln(x) converge. Therefore, I decided that it converges for all values p>1. I'm really not sure if I worked this one correctly or not. Question 2: I worked this one very similarly, but because it seems to be more complex, stumbled. Once again, I took the sister function: f(x)= lnx/x^p integrated by parts: ∫(lnx/x^p)dx u=lnx du=(1/x)dx dv=(1/x^p)dx v=x^(1-p)/(1-p) With this, I'm not completely sure I got the correct value for v, when integrating dv. However, I proceeded to set up the equation: uv - ∫vdu = ((lnx)(x^1-p))/(1-p) - ∫[(x^(1-p))/(x(1-p))]dx By this point, I didn't even bother integrating the ∫vdu because I felt so far off track. That's pretty much as far as I worked on that one, assuming any of it is correct. I'm new to this forum, so I hope I posted all of the necessary information... It looks a bit messy, I tried to tidy it up as well as I could with symbols, but I'm not very familiar with how to use the fractions. Thanks in advance for the help!!!