# 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?

• The change of variable $x=e^t$ and the inequality $e^t>t$ yield $$\int\frac{dx}{\ln(x)^2}=\int\frac{e^tdt}{t^2}>\int\frac{dt}t,$$ from which one can conclude.
– Did
May 4 '16 at 6:15

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.

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!

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.

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

• I'll grant that $\ln(n) < n,$ but how does one prove that $(\ln n)^2 < n$? May 4 '16 at 5:15
• @treble If you know that $\ln(n)<n$, then $\ln(\sqrt{n}) < \sqrt{n}$ and hence $\frac12\ln(n)<\sqrt{n}$, hence $(\ln(n))^2<4n$. May 4 '16 at 6:17
• Regardless it only needs to hold asymptotically, which is quite easy to show. May 5 '16 at 1:07

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.