# For what values of p does

1. Sep 8, 2009

### Hyrox

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!!!

2. Sep 8, 2009

### lanedance

Re: for what values of p does....

i think you mean lim u goes to infinity, which is ok as lnx goes to infinity a s x does

to write in latex, trying clicking on text below to see the script
$$\frac{u^{1-p}}{1-p}$$

you reasoning looks ok to me, but take a look at the first term in your series... may complicate things a little
I think you're on the right track here, try simplifying the integral & keep going

It should also be easy to do a comparision test for p<=1, with the harmonic series 1/n

3. Sep 8, 2009

### Hyrox

Re: for what values of p does....

Okay, let's see.

I stopped here:
uv - ∫vdu = ((lnx)(x^1-p))/(1-p) - ∫[(x^(1-p))/(x(1-p))]dx
and then I picked back up.

$$\frac{lnx(x)^{1-p}}{1-p}$$ -$$\frac{1}{1-p}$$ ∫$$x^(-p)$$

= lim n→∞ $$\frac{ln(x)^{1-p}}{1-p} - \frac{x^{1-p}}{(1-p)^{2}}$$

At this point, I said that the limit converges for all values p > 1, because as p grows past 1, the x will be going to negative exponential powers, which converge to 0. Did I follow through well enough?

P.S. I hope I used latex okay.

Last edited: Sep 9, 2009
4. Sep 9, 2009

### lanedance

Re: for what values of p does....

Another point, for any cases where p = 1, the integration steps will be different so you should consider this separately

Back to Q2, you get to:
$$\displaystyle\lim_{x\to\infty} \frac{ln(x).x^{1-p}}{1-p} - \frac{x^{1-p}}{(1-p)^{2}}$$

I think you dropped the x somewhere previously, but you will need to consider the limit of ln(x)/x^(1-p), for p > 1