Determine whether the series $\sum_{n=2}^{\infty} \frac{n}{4(\ln n)^2}$ converge

Question

I've tried to apply integral test to this series,

$$ \sum_{n=2}^{\infty} \ \frac{n}{4(\ln n)^2} $$

which doesn't work since the result would be a special function. Can you give me a hint on this?

Answer 1

We see that $$\lim _ { x \rightarrow \infty } \left( \frac { x } { 4 \ln \left( x ^ { 2 } \right) } \right) = \infty$$ $$\lim _ { x \rightarrow \infty } \left( \frac { x } { 4 \ln \left( x ^ { 2 } \right) } \right) = \frac { 1 } { 4 } \cdot \lim _ { x \rightarrow \infty } \left( \frac { x } { \ln \left( x ^ { 2 } \right) } \right)$$ We can apply L'Hopital's Rule $$= \frac { 1 } { 4 } \cdot \lim x \rightarrow \infty \left( \frac{1}{\frac{2}{x}} \right)$$ We simplify to get $$= \frac { 1 } { 4 } \cdot \lim x \rightarrow \infty \left( \frac { x } { 2 } \right)$$ Which goes to infinity as $x$ goes to infinity. Therefore, this series does not converge.

Answer 2

As noticed since $a_n \not \to 0$ the series diverges or as an alternative note that since eventually $\sqrt n \ge \log n$ we have

$$\frac{n}{4(\ln n)^2} \ge \frac{1}{4(\ln n)^2} \ge \frac14\frac{1}{n}$$