sequences and series - Is $sum_{n=3}^inftyfrac{1}{nlog n}$ absolutely convergent, conditionally convergent or divergent?

Classify
$$\sum_{n=3}^\infty \frac{1}{n\log(n)}$$
as absolutely convergent, conditionally convergent or divergent.

Is it,
$$\sum_{n=3}^\infty \frac{1}n$$ is a divergent $p$-series as $p=1$, and

$$\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0$$ by comparison test. And this converges to $0$.

So,
$$\sum_{n=3}^\infty \frac{1}{n\log (n)}$$
is conditionally convergence?

I'm not sure if I'm doing right or not. Could you guide me?

Thanks in advance! :)

Answer

Note that if your given series is convergent then it's also absolutely convergent since $\frac{1}{n\log n}\geq 0\quad \forall n\geq 3$.

Now, since the sequence $(\frac{1}{n\log n})$ is decreasing to $0$ so by the integral test your series has the same nature that the improper integral
$$\int_3^\infty\frac{dx}{x\log x}=\left[\log(\log(x))\right]_3^{\to\infty}=\infty$$
hence the series is divergent.