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.


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.

  • 1
    $\begingroup$ $\log (\log 2^n) = \log (n\log 2) = \log n + \log (\log 2)$. Ignore the first few terms maybe, then you have $\frac{1}{2}\log n < \log (\log 2^n) < \log n$ (since $\log (\log 2) < 0$). $\endgroup$ Oct 30, 2014 at 22:01
  • $\begingroup$ @DanielFischer This really works! Thank you very much. $\endgroup$
    – Henry Choi
    Aug 10, 2020 at 9:04

2 Answers 2


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 $p>1$.

  • 1
    $\begingroup$ "$\frac{1}{n(logn+loglog2)^p} \leq \frac{1}{n(logn)^p}$" I don't think this is true, since if you let $n=10$, $p=1.1$, then $\frac{1}{n(logn+loglog2)^p} \approx 0.048 $ and $\frac{1}{n(logn)^p} \approx 0.040 $ $\endgroup$
    – ignoramus
    Nov 1, 2014 at 8:51

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 $$\sum\limits_{n=u}^{v}\frac{1}{n(\ln n)\ldots\left(\ln^{[t-1]}n\right)}\geqslant\int\limits_{u+1}^{v}\frac{dx}{x(\ln x)\ldots\left(\ln^{[t-1]}x\right)}.$$

  • $\begingroup$ This looks very interesting, but I'm having trouble understanding it. Could you explain how the results follow from your two statements? Thanks $\endgroup$
    – ignoramus
    Nov 1, 2014 at 10:24
  • 1
    $\begingroup$ @ignoramus I added the details. $\endgroup$
    – user2097
    Nov 1, 2014 at 11:39
  • $\begingroup$ Ah ok I see, thanks! $\endgroup$
    – ignoramus
    Nov 1, 2014 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.