# $\sum_{n=1}^{\infty}\frac{1}{n(\ln{n})^2+n}$

Does the following series converge or diverge $$\sum_{n=1}^{\infty}\frac{1}{n(\ln{n})^2+n}$$ I know that $\sum_{n=1}^{\infty}\frac{1}{(\ln{n})^2}$ diverges.

$\sum_{n=1}^{\infty}\frac{1}{n(\ln{n})^2}$ dominates the series in question, and converges. What comparison can I use to understand this?

• Useful keyword: it's a Bertrand series. – Clement C. Jun 20 '16 at 15:34
• Show that the convergence of your series is implied by the convergence of $\sum_{n=2}^{\infty} \frac 1{n(\ln n)^2}$. Then use integral test. – Sungjin Kim Jun 20 '16 at 16:05

$$\frac{1}{n\log^2(n)+n}\le \frac{1}{n\log^2(n)}$$
$$\int_2^\infty \frac{1}{x\log^2(x)}\,dx=\frac{1}{\log(2)}$$
The Cauchy condensation test gives that your series is convergent iff $$\sum_{m\geq 0}\frac{1}{1+m^2}=\frac{1+\pi\coth \pi}{2}$$ is convergent.