Convergence of $\sum\limits_{n=2}^{\infty} \frac{1}{\ln(n)^2}$

Question

I am trying to use the integral test on the series $$\sum\limits_{n=2}^{\infty} \frac{1}{\ln(n)^2}.$$ I am not sure how to evaluate the integral. Any hints?

Answer 1

In order to use the integral test, you need to figure out whether or not the integral $$\int_2^{\infty} \frac{1}{(\ln x)^2} dx $$ converges or not. To do this, I would suggest making the substitution $u = \ln (x),$ then note that $du = 1/x dx$ and that $x = e^u$. Plug all of this in and go to work!

Answer 2

The sum is clearly bounded below by the harmonic series starting from 2 (since $\ln(n)^2 < n$), so it diverges.

Answer 3

One may onserve that $$ 0<\ln x<\sqrt{x} ,\qquad x>2, $$ giving $$ 0<(\ln x)^2<x, \qquad x>2, $$ then

$$ \int_2^M \frac{dx}x < \int_2^M \frac{dx}{(\ln x)^2}, \qquad M>2, $$

thus, by letting $M \to \infty$, the initial integral diverges.

Answer 4

Regarding

\begin{equation} \sum_{n=2}^{\infty}\frac{1}{\left(\ln x\right)^2} \end{equation}

you should be able to show that $f(x)=\sqrt{x}-\ln x$ is positive for $x\ge2$. (Just find the minimum value of $f(x)$.)

Therefore for $n\ge2$ it's true that $0<\ln n<\sqrt{n}$.

So $\dfrac{1}{(\ln{n})^2}>\dfrac{1}{n}$

Therefore the series $\sum_{n=2}^{\infty}\frac{1}{\left(\ln x\right)^2}$ diverges by direct comparison to the harmonic series.