Rudin 3.29 the series $\sum_{n=3}^{\infty} \frac{1}{n \ln n(\ln(\ln n))^2}$ converges

Question

In Rudin's Principles of Mathematical Analysis, we claim (with no proof of it) that $$\sum_{n=3}^{\infty} \frac{1}{n \ln n(\ln(\ln n))^2}$$ is convergent.

After applying the Cauchy compression test (the term of the series is a decreasing, positive sequence) I get: $$\sum_{n=3}^{\infty} \frac{1}{n \ln n(\ln(\ln n))^2} = \frac{1}{\ln 2} \sum _{n=2}^{\infty} \frac{1}{n} \frac{1}{(\ln n + \ln (\ln 2))^2}$$

• Do we have to use a certain majoration of the term of the series and conclude by the comparison test ? (since $\ln (\ln 2)< 0$ it seems quite impossible)

• Do we have to use these considerations ? (they seem a bit too difficult to understand for me)

• Or is there a $3^{rd}$ way to prove the convergence ?

Answer 1

I figured out my mistake.

Consider the integral of this expression:

$$\int_3^\infty\frac{dn}{n\ln(n)\cdot(\ln(\ln(n))^2}$$

Note the following:

$$\frac{d}{dn}\ln(\ln(n))=\frac{1}{n\ln(n)}$$

$$\implies I=\int_{\ln(\ln(3))}^\infty\frac{du}{u^2}$$

$$\implies I=\left[-\frac{1}{u}\right]_{\ln(\ln(3))}^\infty$$

$$=\frac{1}{\ln(\ln(3))}$$

So the sum converges by the integral series test

Answer 2

We will prove the convergence of the OP’s series by applying the Cauchy compression test twice.

$\begin{align}\sum_{n=3}^{\infty}\frac1{n\ln n\big(\!\ln(\ln n)\big)^{\!2}}&\leqslant\sum_{n=2}^{\infty}\frac{2^n}{2^n\ln\left(2^n\right)\big(\!\ln(\ln(2^n))\big)^{\!2}}=\\[2pt]&=\sum_{n=2}^{\infty}\frac1{n\ln2\big(\!\ln n+\ln(\ln2)\big)^{\!2}}\leqslant\\[2pt]&\leqslant\sum_{n=1}^{\infty}\frac{2^n}{2^n\ln2\big(\!\ln(2^n)+\ln(\ln2)\big)^{\!2}}=\\[2pt]&=\sum_{n=1}^{\infty}\frac1{\ln2\big(n\ln2+\ln(\ln2)\big)^{\!2}}<\\[2pt]&<\sum_{n=1}^{\infty}\frac1{\ln2\,\left(\frac n4\right)^2}=\dfrac{16}{\ln2}\sum_{n=1}^{\infty}\frac1{n^2}=\\[2pt]&=\dfrac{16}{\ln2}\cdot\dfrac{\pi^2}6=\dfrac{8\pi^2}{3\ln2}\,.\end{align}$

Answer 3

Just to expand on @Angelo's answer and generalize, there is no need to explicitly calculate any bounds on this sum. This may also make it easier to see what Rudin means when he says "if we continue this process..."

First consider $p > 0$,

Then $\frac{1}{n \ln n (\ln \ln n)^p}$ is a monotoincally decreasing, non negative sequence for $n \geq 3$, so by 3.27,

$\sum_{n=3}^{\infty} \frac{1}{n \ln n(\ln \ln n)^p}$ converges if and only if $\sum_{n=2}^{\infty} \frac{2^{n}}{2^{n} n\ln 2(\ln n \ln 2)^p}$ = $\frac{1}{\ln 2} \sum_{n=2}^{\infty} \frac{1}{ n(\ln( n \ln 2))^p}$ converges.

Now $ \frac{1}{ n(\ln( n \ln 2))^p}$ is a monotonically decreasing, non negative sequence for $n\geq 2$, so by 3.27,

$\frac{1}{\ln 2} \sum_{n=2}^{\infty} \frac{1}{ n(\ln( n \ln 2))^p}$ converges if and only if

$\frac{1}{\ln 2} \sum_{n=1}^{\infty} \frac{2^n}{ 2^n (\ln( 2^n \ln 2))^p} = \frac{1}{\ln 2 \ln(2 \ln 2)^p} \sum_{n=1}^{\infty} \frac{1}{ n^p}$ converges. Which by 3.28 converges if $p > 1$ and diverges if$ p \leq 1$.

For $p \leq 0$, $\frac{1}{n \ln n(\ln \ln n)^p} \geq \frac{1}{n \ln n} \geq 0$ for $n \geq 3$, so by 3.29 and 3.25 b, we have that $\sum_{n=3}^{\infty} \frac{1}{n \ln n(\ln \ln n)^p} $ diverges for $p \leq 0$.

Thus $\sum_{n=3}^{\infty} \frac{1}{n \ln n(\ln \ln n)^p}$ diverges for $p \leq 1$ and converges for $p > 1$.