Help with Baby Rudin Theorem 3.29

Question

Theorem 3.29 If $p >1$ then $\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p}$ converges; if $p \leq 1$, the series diverges.

Proof:

The monotonicity of the logarithmic function implies that $\{\log n\}$ increases. Hence $\frac{1}{n\log n}$ decreases, and we can apply theorem 3.27; this leads us to the series

$\sum_{k=1}^{\infty} 2^k \frac{1}{2^k(\log 2^k)^p}= \sum_{k=1}^{\infty} \frac{1}{(k\log 2)^p}= \frac{1}{(\log 2)^p} \sum_{k=1}^{\infty} \frac{1}{k^p}$ and the conclusion follows from Theorem 3.28.

I understand the proof thus far, but Rudin goes on to say that this procedure may evidently be continued. For instance,

$\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)}$ diverges, whereas

$\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^2}$ converges.

I want to continue the procedure to show for which $p$ does $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converge.

Here are the relevant theorems:

Theorem 3.27 Suppose $a_1 \geq a_2 \geq ... \geq 0$. Then the series $\sum_{k=1}^{\infty} a_n$ converges if and only if $\sum_{k=1}^{\infty} 2^n a_{2^n}$ converges.

Theorem 3.28 $\sum \frac{1}{n^p}$ converges if $p>1$ and diverges if $p \leq 1$.

Here is what I have so far:

By Theorem 3.27 $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converges if and only if $\sum_{k=2}^{\infty}2^n \frac{1}{2^n\log 2^n(\log\log 2^n)^p} = \frac{1}{\log 2} \sum_{k=2}^{\infty} \frac{1}{n(\log\log 2^n)^p}$

Now I'm stuck. How may the procedure be continued? Please note it is my goal to show a result similar to Theorem 3.29. It is not my goal to simply show that $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converges for some $p$. Any hint would be greatly appreciated. Thank you.

Answer 1

I think I figured it out.

By Theorem 3.29 if $p>1$, then $\Sigma_{n=2}^{\infty}\frac{1}{n(logn)^p}$ converges.

By Theorem 3.27 $\Sigma_{n=3}^{\infty} \frac{1}{nlogn(loglogn)^p}$ converges if and only if $\frac{1}{log2} \Sigma_{n=2}^{\infty} \frac{1}{n(loglog2^n)^p}$ converges.

But $\frac{1}{n(loglog2^n)^p} = \frac{1}{n(logn+loglog2)^p} \leq \frac{1}{n(logn)^p}$

Hence $\frac{1}{log2} \Sigma_{n=2}^{\infty} \frac{1}{n(loglog2^n)^p}$ converges for $p>1$, and therefore $\Sigma_{n=3}^{\infty} \frac{1}{nlogn(loglogn)^p}$ converges for $p>1$.

Answer 2

Denoting by $\ln^{[t]}$ the $t$th iteration of logarithm, one can get $$\frac{1}{\ln^{[t]} n}-\frac{1}{\ln^{[t]} (n+1)}=\frac{1}{n(\ln n)\ldots\left(\ln^{[t]}n\right)\left(\ln^{[t+1]}n\right)^2}+O\left(\frac{1}{n^2}\right),$$ so that $$\sum\limits_{n=N}^\infty \frac{1}{n(\ln n)\ldots\left(\ln^{[t]}n\right)\left(\ln^{[t+1]}n\right)^2}\leqslant\sum\limits_{n=N}^\infty\left(\frac{1}{\ln^{[t]} n}-\frac{1}{\ln^{[t]} (n+1)}\right)+c=\frac{1}{\ln^{[t]} N}+c,$$ which proves the first part of the result. The other part is done just by noting that $$\frac{d\,\ln^{[t]}x}{dx}=\frac{1}{x(\ln x)\ldots\left(\ln^{[t-1]}x\right)},$$ because $$\sum\limits_{n=u}^{v}\frac{1}{n(\ln n)\ldots\left(\ln^{[t-1]}n\right)}\geqslant\int\limits_{u+1}^{v}\frac{dx}{x(\ln x)\ldots\left(\ln^{[t-1]}x\right)}.$$