# Divergence ~ ln(ln(ln x))

• October 18th 2009, 12:48 PM
redsoxfan325
Divergence ~ ln(ln(ln x))
It is well-known in high-level mathematics that:

$\lim_{n\to\infty}\left[\left(\sum_{k=1}^n\frac{1}{k}\right)-\ln(n)\right]=\gamma$ (Euler-Mascheroni Constant)

$\lim_{n\to\infty}\left[\left(\sum_{p~prime}^n\frac{1}{p}\right)-\ln(\ln(n))\right]=M$ (Meissel-Mertens Constant)

Is there a divergent series $\sum a_k$ such that:

$\lim_{n\to\infty}\left[\left(\sum_{k=1}^n a_k\right)-\ln(\ln(\ln(n)))\right]=constant$

If so, can it be generalized to $\ln(\ln(...(\ln(n))))$?
• October 19th 2009, 12:51 AM
Opalg
The approximation $\sum_{k=1}^n\frac{1}{k}\approx\ln(n)$ can be justified by applying the procedure of the integral test for convergence to get $\sum_{k=1}^n\frac{1}{k}\approx\int_1^n\frac1x\,dx = \ln(n)$.

A similar procedure will give you $\sum_{k=1}^n\frac{1}{k\ln k}\approx\ln(\ln(n))$, $\sum_{k=1}^n\frac{1}{k\ln(\ln k)}\approx\ln(\ln(\ln n)))$, $\sum_{k=1}^n\frac{1}{k\ln(\ln(\ln k))}\approx\ln(\ln(\ln(\ln n))))$, and so on.

Edit. See corrections below.
• October 19th 2009, 01:26 AM
chisigma
Is...

$\frac{d}{dx} \ln x = \frac{1}{x} \rightarrow \int \frac{dx}{x} = \ln x + c$ (1)

... all right!...

Is...

$\frac{d}{dx} \ln (\ln x) = \frac{1}{x\cdot \ln x} \rightarrow \int \frac{dx}{x\cdot \ln x} = \ln (\ln x) + c$ (2)

... all right!...

But is...

$\frac{d}{dx} \ln (\ln (\ln x))= \frac{1}{x\cdot \ln x \cdot \ln (\ln x)} \rightarrow \int\frac {dx}{x\cdot \ln(\ln x)} \ne \ln (\ln (\ln x))+c$ (3)

... something wrong from me?(Thinking) ...

Kind regards

$\chi$ $\sigma$
• October 19th 2009, 02:17 AM
Opalg
Quote:

Originally Posted by chisigma
But is...

$\frac{d}{dx} \ln (\ln (\ln x))= \frac{1}{x\cdot \ln x \cdot \ln (\ln x)} \rightarrow \int\frac {dx}{x\cdot \ln(\ln x)} \ne \ln (\ln (\ln x))+c$ (3)

... something wrong from me?(Thinking) ...

You're right, of course. The series should be $\sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}$. (While we're about it, the series has to start at k=2, not k=1, for obvious reasons.)

Similarly, the series that converges at the rate of ln(ln(ln(ln n))) should have been $\sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}$, and so on.
• October 19th 2009, 09:43 AM
redsoxfan325
Quote:

Originally Posted by Opalg
You're right, of course. The series should be $\sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}$. (While we're about it, the series has to start at k=2, not k=1, for obvious reasons.)

Similarly, the series that converges at the rate of ln(ln(ln(ln n))) should have been $\sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}$, and so on.

Is there a meaningful infinite set $A\subset\mathbb{N}$, such that

$\sum_{k\in A}\frac{1}{k}$ diverges ~ $\ln(\ln(\ln n)))$

Like maybe let $A=\{p\in\mathbb{N}:p~prime,~p\equiv1\mod4\}$

But, for all I know, that could converge. (I know that if $A=\{p:p~prime,~p+2~also~prime\}$, then the series does converge.)

Actually, now that I think about it, that sum must converge, because

$\sum_{p\in A}\frac{1}{p}=\sum_{m,n\in I\subset\mathbb{N}}\frac{1}{m^2+n^2}<\sum_{k=1}^{\ infty}\frac{1}{k^2}=\frac{\pi^2}{6}$

EDIT: Although, the above calculation seems to imply that if we let $B=\{p:p~prime,~p\equiv3\mod4\}$ then $\sum_{p\in B}\frac{1}{p}$ diverges.
• October 21st 2009, 08:58 PM
chiph588@
$\sum_{p \equiv 1 \; \text{mod} \; 4} \frac{1}{p}$ diverges because one can show that given $(a,k)=1$ and $A = \{$ prime $p \; | \; p \equiv a \; \text{mod} \; k \}$, $\lim_{s \to 1^{+}} \frac{\sum_{p \in A} \frac{1}{p^s}}{\sum_{p} \frac{1}{p^s}} = \frac{1}{\phi(k)}$.
• October 21st 2009, 09:01 PM
redsoxfan325
What's $\phi(k)$? (And also, where does my proof of convergence go wrong?)
• October 21st 2009, 09:06 PM
chiph588@
Quote:

Originally Posted by redsoxfan325
What's $\phi(k)$? (And also, where does my proof of convergence go wrong?)

$\phi(k)$ is Euler's totient function. In your proof, What is $I$?
• October 21st 2009, 09:11 PM
redsoxfan325
Quote:

Originally Posted by chiph588@
$\phi(k)$ is Euler's totient function. In your proof, What is $I$?

If $p\equiv1\mod4$, then $p=m^2+n^2$ for some $m,n\in\mathbb{N}$. $I$ is the subset of $\mathbb{N}$ that contains all of those $m$ and $n$. It might be all of $\mathbb{N}$, but I wasn't sure, and I could think of a good way to write it, so I wrote $I$.
• October 21st 2009, 09:25 PM
chiph588@
I'd say this equality is incorrect:
$\sum_{p \in A} \frac{1}{p} = \sum_{m,n\in I\subset\mathbb{N}}\frac{1}{m^2+n^2}$. This is because for a fixed $m$ (or $n$), the latter sum doesn't take into account how many times the $m$ (or $n$) appear in forming an arbitrary $p$.