Divergent or not series?

Question

How tipe are this series? $$\sum_{n=2}^\infty \frac 1 {n^2\ln n}$$ But $$\sum_{n=2}^\infty \frac 1 {n\ln n\ln(\ln n)}$$ I used Condensation Test, but I stack after that. Thanks!

Answer 1

Let $f(n)=1/n^2\ln n$ for $n\geq 2$. Then $f(n)$ is decreasing. Now $0<2^nf(2^n)=1/(2^n\cdot n\ln 2)<1/2^n,$ and $\sum_{n>1}1/2^n$ converges, so $\sum_{n>1} f(n)$ converges.

Let $g(n)=[n(\ln n)(\ln \ln n)]^{-1}$ for $n\geq 2.$ Then $g(n)$ is decreasing for $n\geq 3.$ And $$2^n g(2^n)= [(n\ln 2)(\log n +\ln \ln 2)]^{-1}>[2\ln 2]^{-1}[n\ln n]^{-1}>0$$ for $n\geq 3.$ Applying the Cauchy condensation test to $[n\ln n]^{-1}$ shows that $\sum_{n>2}[n\ln n]^{-1}$ diverges, so $\sum_{n>1}g(n)$ diverges.

Note: It often helps to look for ways to simplify an intermediate step. E.g. by comparing $2^ng(n)$ to $[2\ln 2]^{-1}[n\ln n]^{-1} ,$ as there is less calculation and detail when applying the condensation test to the latter expression.

Note: By repeated use of the condensation test, each of the series $\sum_{n>n_1} [n(\ln n)^A]^{-1},\; \;$ $\sum_{n>n_2} [n(\ln n) (\ln \ln n)^A]^{-1},\;\; $ $\sum_{n>n_3} [n(\ln n)( \ln \ln n) (\ln\ln\ln n)^A]^{-1}$...(et cetera)... (where $n_1,n_2,n_3$ are large enough that the terms of the series are all defined).... converges for $A>1$ and diverges for $A\leq 1.$

Answer 2

Hint

$$\frac{1}{n^2 \ln(n)} <\frac{1}{n^2} $$

For the second one, calculate $\int_2^R \frac{1}{x \ln (x) \ln(\ln(x))}$ using the substitution $u=\ln(\ln(x))$.