What would you like to ask?what values of p the series ∑ln(n)/n^p is convergent

I get stuck at lim t--> infinity [ln(n)t^1-p / 1-p - t^1-p/ (1-p)^2] from 1 to t ........
I took the integral for lnn/n^p and set the interval [1, infinity)
t substitutes for infinity. I don't know what else i should do. I know i have to take the integrals and show that it goes to 0. This is the way p>1 can be proven correct and that it converges.
Please help! :(. Can a step by step process be given?

For p > 1:
∫(1 to ∞) ln(x) dx / x^p
= lim(t→∞) ∫(1 to t) ln(x) dx / x^p
= lim(t→∞) [(ln x) x^(-p+1)/(-p+1) {for t = 1 to t} - ∫(1 to t) x^(-p) dx/(-p+1)]
= lim(t→∞) [(ln x) x^(-p+1)/(-p+1) - x^(-p+1)/(-p+1)^2] {for t = 1 to t}
= lim(t→∞) [(ln t) t^(-p+1)/(-p+1) - t^(-p+1)/(-p+1)^2] - [0 - 1/(-p+1)^2]
= 1/(-p+1)^2 + (1/(-p+1)^2) * lim(t→∞) [(-p+1) (ln t) - 1] / t^(p-1)
= 1/(-p+1)^2 + (1/(-p+1)^2) * lim(t→∞) [(-p+1) (1/t)] / [(p-1) t^(p-2)], by L'Hopital's Rule
= 1/(-p+1)^2 + (1/(-p+1)^2) * lim(t→∞) -1/t^(p-1)
= 1/(-p+1)^2 + 0
= 1/(p-1)^2.
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Just in case:

For p = 1:
∫(1 to ∞) ln(x) dx / x = (1/2) (ln x)^2 {for x = 1 to ∞} = ∞

For p < 1:
Using the work above, ∫(1 to ∞) ln(x) dx / x^p
= lim(t→∞) [(ln x) x^(-p+1)/(-p+1) - x^(-p+1)/(-p+1)^2] {for t = 1 to t}
= lim(t→∞) [(ln t) t^(-p+1)/(-p+1) - t^(-p+1)/(-p+1)^2] - [0 - 1/(-p+1)^2]
= 1/(-p+1)^2 + (1/(-p+1)^2) * lim(t→∞) [(-p+1) (ln t) - 1] / t^(p-1)
= 1/(-p+1)^2 + (1/(-p+1)^2) * lim(t→∞) [(-p+1) (ln t) + 1] t^(1-p)
= ∞.

So, the integral only converges for p > 1.

I hope this helps!

Did you use partial fractions? I get confused in some places.

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Yes, you're right; any time a sum/integral yields a finite answer, it converges.

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holy **** math is getting complicated. make an argument for the overall cap loimit far below infinte that can be expressed in the near future in 3d media devices.