Integral Test for Convergence - Log of a Log

I need to investigate the convergence of the series $\sum_{n=3}^{\infty}\dfrac{1}{n\ln n}$

So, doing the integral test, I end up with (shortcutted because integration is boring):

$\lim_{M\rightarrow\infty}\int_{3}^{M}\dfrac{1}{x\ln x}dx=\ldots=\lim_{M\rightarrow\infty}\left[\ln(\ln M))-\ln(\ln3))\right]+c$

But as $M \rightarrow \infty$, this value will also tend to infinity.

I've been asked to then investigate the convergence of the general form: $\sum_{n=3}^{\infty}\dfrac{1}{n\ln^{q} n}$, $q \geq 1$, but if it does not converge for $q=1$, then I doubt it will for any higher values... and since this is a follow up question, it makes me believe that I've missed something somewhere, or there is more here than what I'm seeing.

Is the series divergent as I think? Or is it convergent and my integration is wrong?

• What does the Integral Test tell you? – David Mitra Jun 4 '14 at 11:14
• The series (for $q = 1$) is divergent, your integration is correct. – M. Vinay Jun 4 '14 at 11:15
• Don't assume it diverges for higher values also, there's no way you could conclude this. For example, $\sum\dfrac{1}{n^p}$ is divergent for $p = 1$, but convergent for $p > 1$. So don't jump to conclusions, do the integral test for $q > 1$. – M. Vinay Jun 4 '14 at 11:18
• Thanks! I'm always wary when two questions ask me to prove something, but they both end up having the same properties or something to this effect. I guess I'm used to "showing convergence" rather than "proving divergence". – Yoshi Jun 4 '14 at 11:19
• don't forget the second requirement for the integral test: the function must be decreasing, otherwise you may find a diverging sum for which the corresponding integral converges – Alex Jun 4 '14 at 11:21

$$\int_3^M \frac{dx}{x \log^q{x}} = \int_3^M \frac{d(\log{x})}{\log^q{x}} = \frac1{q-1} \left (\frac1{\log^q{3}} - \frac1{\log^q{M}} \right )$$
For $q \gt 1$, this converges in the limit as $M \to \infty$.
• Thanks! I really have to watch those assumptions, it was only once I sat down to actually work out the case for $q \geq 2$ that I realized it would yield a very different answer to the case for $q=1$. – Yoshi Jun 4 '14 at 12:02
Condensation Test ($\;q>1$):
$$\frac{2^n}{2^n\log^q(2^n)}=\frac1{n^q\log 2}$$