Convergence of series 4 Determine if the following series is convergent or not:
$$\frac{1}{\sqrt{n} \log n}$$
I tried:
$a_k = \frac{1}{\sqrt{n} \log n}$
$b_k= \frac{1}{\sqrt{n}}$
then did: $\frac{a_k}{b_k}$
and got $\frac{1}{\log n}$.
 A: I'm assuming you aim to determine the convergence of $$\sum_{n=2}^\infty \frac{1}{\sqrt{n}\log(n)}$$ Let's try to find a series that is smaller than this one, but still divergent. For example, $$\sum_{n=2}^\infty \frac{1}{n\log(n)}$$ By the integral test we have $$\int_{2}^\infty \frac{1}{x\log(x)}\text{d}x = \log(\log(x))\Big\vert^\infty_2 = \infty $$ so $$\sum_{n=2}^\infty \frac{1}{n\log(n)}$$ diverges, which means the larger series in question must be divergent.
A: One good option here is the logarithmic test: compute $\lim \limits _{n \to \infty} \frac {\log \frac 1 {x_n}} {\log n}$. If you get a number $<1$, the series is divergent; if you get $>1$, it is convergent; if you get $1$, you cannot draw any conclusion and you must try some other method. The test works only for series with positive terms.
In your case, $\lim \limits _{n \to \infty} \frac {\log (\sqrt n \log n)} {\log n} = \lim \limits _{n \to \infty} \frac {\frac 1 2 \log n + \log \log n} {\log n} = \frac 1 2 < 1$, so the series is divergent.
Another option would have been to use Cauchy's condensation test.
