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

Question

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?

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
then it should have been written as $(\ln n)^2$ or $\ln^2n$, not $\ln(n)^2$. by putting brackets around $n$, it seems the "square" only applies to it. — , Nov 18 '12 at 22:12
Note that, as R.J. Mathar posted in his not-an-answer answer, «The numerical constant and further literature are in http://oeis.org/A115563 .». — MickG, Dec 03 '15 at 20:47
how do I find an integer such that $$\left\lvert s-\sum_{n=2}^\infty \frac{1}{n;(ln;n)^2}\right\rvert < 10^{-3}.$$ where $s=\sum_{n=2}^\infty \frac{1}{n;(ln;n)^2}$ — stackdsewew, Jun 12 '16 at 11:20

score 10 · Accepted Answer · answered Nov 18 '12 at 21:52

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.

score 8 · Answer 2 · edited Dec 03 '15 at 20:33

8

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.

edited Dec 03 '15 at 20:33

user26857

52,094

answered Nov 18 '12 at 22:04

P..

14,929

1

Thank you -- this is what I ended up using, but I decided to accept Marvis's answer because it helped show me why the limit test wasn't working here. – Michael0x2a Nov 18 '12 at 22:12
The integral converges to 2, but is it correct to say that the series also converges to 2, or can we not make that conclusion? – stackdsewew Jun 12 '16 at 09:09
@anna_xox I found that it converges to $\frac{1}{ln(2)}$ - did I calculate wrong? – Elad Avron Jun 27 '16 at 10:17

score 6 · Answer 3 · edited Dec 03 '15 at 20:35

6

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.

edited Dec 03 '15 at 20:35

user26857

52,094

answered Nov 18 '12 at 21:56

Dennis Gulko

15,640

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

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