Whether this series converge or not.

Question

\begin{equation} a_n = \frac{1}{n(\log n)^p(\log \log n)^q}, \text{where } p,q \ge 0. \end{equation} Does $\sum_{n=2}^\infty a_n$ converge? Should I use Bertrand's method?

Answer 1

From Cauchy's Condensation test the the series $\sum_{n=2}^\infty \frac{1}{n(\log(n))^p(\log(\log(n)))^q}$ converges or diverges if and only if the condensed series, given by

$$\sum_{n=2}^\infty \frac{2^n}{(2^n)\,(\log(2^n))^p(\log(\log(2^n))^q}=\frac{1}{\log^p(2)}\sum_{n=2}^\infty \frac{1}{n^p(\log(n\log(2))^q} \tag 1$$

converges or diverges.

Now, using Bertrand's Methodology, we see that the series on the right-hand side of $(1)$ diverges for $p<1$, converges for $p>1$, and for $p=1$ the series converges for $q>1$ and diverges otherwise.

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Alternatively, by enforcing the substitution $x=e^t$, we can see that

$$\int_2^\infty \frac{1}{x(\log(x))^p(\log(\log(x)))^q}\,dx=\int_{\log(2)}^\infty \frac{1}{t^p\log^q(t)}\,dt$$

which converges for $p>1$, diverges for $p<1$, and for $p=1$, converges for $q>1$ and diverges otherwise, which agrees with the previous result.

Answer 2

I assume you mean:

When does $\sum a_n$ converges regarding $p$ and $q$?

Hint

Use a comparaison to an integral to evaluate the convergence.

(This is indeed similar to Bertrand's method)