Why does $\sum_{n=2}^ \infty \frac 1 {n \sqrt {\log n}}$ diverge?

Question

I'm aware that this diverges, but I'm struggling to prove it only using a comparison, or a ratio/root test. Any tips would be greatly appreciated - this isn't even an assignment question, just one that I can't seem to get out of my head.

Answer 1

A good observation here is the series diverges if the following indefinite integral diverges: $$ \int_2^\infty \frac{1}{x \sqrt{\ln x}} dx $$ Notice how I used $\ln x$ instead of $\log x$? The base change does not affect the evaluation in this particular situation, it just makes it easier to compute the integral. We are going to use substitution: Let $u = \ln x$ then $du = \frac1x dx \Leftrightarrow x du= dx$. We have the following: $$ \int_{\ln (2)}^\infty \frac{du}{\sqrt{u}} = (2\sqrt u) |^\infty_{\ln 2} = \infty $$

The integral diverges, so the series diverges.

Answer 2

The sum of 1/n diverges, but just very slowly. Check out why it diverges and why it diverges so slowly.

You added a factor that makes it grow slower. Depending on that factor, it might sill be divergent, but even slower. It might be enough to limit the series. Or it might make a proof difficult. In this case it’s still divergent.

For the sum from 1 to 2, from 3 to 4, from 5 to 8, from 9 to 16, from 17 to 32 and so on find a simple upper and lower limit. So the sum is between xxx and yyy. And then you sum this up. For the sum of 1/n, the sum up to 2^k is at least some constant times k. Do the same here.