Convergence of $\sum\limits_{n=2}^{\infty} \frac{1}{\ln(n)^2}$ 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?
 A: 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.
A: 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!
A: 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.
A: The sum is clearly bounded below by the harmonic series starting from 2 (since $\ln(n)^2 < n$), so it diverges.
A: Use the Cauchy condensation test. Note that 
$$2^na_{2^n}=\frac{2^n}{(\ln 2^n)^2}=\frac{2^n}{n^2(\ln 2)^2}$$so that $\sum 2^n a_{2^n}$ does not converge (terms don't go to zero) and hence the original series diverges.
