# Showing convergence of an infinite series?

I'm trying to show whether the below series converges or diverges, and I have very little clue on how to do it. I know about the Comparison Test, but I can't think of a sequence $$b_n > a_n$$ to perform a comparison with.

Esentially, my question is more how should I approach this problem, and is there a less "guess-based" approach for finding a $$b_n$$? $$\sum_{n=2}^\infty a_n =\sum_{n=2}^\infty\frac{1}{\log(\log(n))\log(n)n}$$

• The Cauchy Condensation test which is based on the comparison test works great for these types of series. Feb 6 '19 at 15:45

Use the integral test. The series you presented is convergence if and only if

$$\int_2^\infty \frac{1}{\log\log(x)\log(x) x} dx <\infty$$

Now substitute $$x=e^x$$ (and remember to multiply by the derivative of $$e^x$$ whihc is $$e^x$$) then we have

$$=\int_{\log(2)}^\infty \frac{1}{\log(x)xe^x}\cdot e^x dx = \int_{\log(2)}^\infty \frac{1}{\log(x)x} dx$$

substitue $$x=e^x$$ once again you will get

$$\int_{\log\log(2)}^\infty \frac{1}{x}dx$$ which diverge.

Note that this trick would work even if you increase the number of $$\log$$. For instance it would also mean that the series diverge for $$a_n = \frac{1}{\log\log\log(n)\log\log(n)\log(n)n}$$ and so on. For each component of $$\log$$ you will need to substitue $$x=e^x$$ one more time.