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! 🙂
Best 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.