# Proving the convergence of $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$

I am attempting to prove if the below expression converges or diverges:

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

I decided to try using the limit convergence test. So then, I set $f(n) = \frac{1}{n (\ln n)^2}$ and $g(n) = \frac{1}{n}$ and did the following:

$$\lim_{n \to \infty} \frac{f'(n)}{g'(n)} = \lim_{n \to \infty} \frac{-\frac{\ln(n) + 2}{n^2 (\ln n)^3}}{-\frac{1}{n^2}} = \lim_{n \to \infty} \frac{\ln(n) + 2}{n^2 (\ln n)^3} \cdot n^2 = \lim_{n \to \infty} \frac{1}{(\ln n)^2} + \frac{2}{(\ln n)^3} = 0$$

I know that $\sum \frac{1}{n}$ diverges due to properties of harmonic series, and so concluded that my first expression $\frac{1}{n (\ln n)^2}$ also must diverge.

However, according to Wolfram Alpha, and as suggested by this math.SE question, the expression actually converges.

What did I do wrong?

• You can use Cauchy's condensation test. – Amihai Zivan Nov 18 '12 at 21:50
• Your $f(x),g(x)$ do not depend on $x$, hence $f'(x)=g'(x)=0$. But you derive them $dn$, where $n$ is discrete. You should take much more care when doing things like that. – Dennis Gulko Nov 18 '12 at 21:52
• The integral test works well too. – mrf Nov 18 '12 at 21:55
• is it $\log(x^2)$ or $(\log x)^2$? – user31280 Nov 18 '12 at 22:04
• @F'OlaYinka -- the latter – Michael0x2a Nov 18 '12 at 22:10

In the limit comparison test, $\sum_n f(n)$ and $\sum_n g(n)$ both will behave the same iff $$\lim_{n \to \infty} \dfrac{f(n)}{g(n)} = c \in (0,\infty)$$ In your case, the limit is $0$ and hence you cannot conclude anything.
You can use integral test for convergence. Since the integral $$\int_2^\infty\dfrac{1}{x\log^2 x}dx$$ converges, the series $\displaystyle{\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}}$ converges.
• @anna_xox I found that it converges to $\frac{1}{ln(2)}$ - did I calculate wrong? – Elad Avron Jun 27 '16 at 10:17
Expanding on the accepted answer, you should use the Cauchy condensation test: For a positive non-increasing sequence $f(n)$, the series $\sum_{n=0}^\infty f(n)$ converges if and only if the series $\sum_{n=0}^\infty2^nf(2^n)$ does.
In your case, $f(n)=\frac{1}{n\ln^2n}$, so $$2^nf(2^n)=2^n\frac{1}{2^n\ln^22^n}=\frac{1}{n^2\ln^22}=\frac{1}{\ln^22}\frac{1}{n^2}$$ which clearly converges.