# Thread: For what Values to the series converge

1. ## For what Values to the series converge

I am struggling with this one. For what values for p does the series converge?

$\sum_{n=1}^\infty$ $ln(n)/n^p$

2. Originally Posted by zifnib
I am struggling with this one. For what values for p does the series converge?

$\sum_{n=1}^\infty$ $ln(n)/n^p$
Hint: $\exists{N}\backepsilon\forall{n\geqslant{N}}~\ln(n )\leqslant{n^{\varphi>0}}$

Or use the Ratio test

3. 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 $ln(n)/n$ 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???

4. Your right about the Ratio test sorry, but why not try the integral test

$\int_1^{\infty}\frac{\ln(n)}{n^p}dn=\lim_{n\to\inf ty}\bigg[\frac{1}{1-p}n\ln(n)-\frac{1}{1-2p+p^2}n\bigg]n^{-p}$

Now find for what values that is convergent.

5. I tried that on my TI-89 but I was not sure to to tell which values for p would work that way? Would I just need to find where the limit is a number other than 0 or infinity? Something seams incorrect to me kinda.

6. Originally Posted by zifnib
I tried that on my TI-89 but I was not sure to to tell which values for p would work that way? Would I just need to find where the limit is a number other than 0 or infinity? Something seams incorrect to me kinda.
Don't use your TI-89! Use your brain! Distribute $n^{-p}$ 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 $\sum_{n=1}^{\infty}2^n\cdot\frac{\ln\left(2^n\righ t)}{2^{np}}=\sum_{n=1}^{\infty}\frac{\ln(2)n}{2^{( p-1)n}}$

Now think what values of p does this serise converges...this time its pretty apparent.