Thread: Divergence ~ ln(ln(ln x))

It is well-known in high-level mathematics that:

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

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

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

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

If so, can it be generalized to $\displaystyle \ln(\ln(...(\ln(n))))$?

The approximation $\displaystyle \sum_{k=1}^n\frac{1}{k}\approx\ln(n)$ can be justified by applying the procedure of the integral test for convergence to get $\displaystyle \sum_{k=1}^n\frac{1}{k}\approx\int_1^n\frac1x\,dx = \ln(n)$.

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

Edit. See corrections below.

Is...

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

... all right!...

Is...

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

... all right!...

But is...

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

... something wrong from me? ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Originally Posted by chisigma

But is...

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

... something wrong from me? ...

You're right, of course. The series should be $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}$. (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 $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}$, and so on.

Originally Posted by Opalg

You're right, of course. The series should be $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)}$. (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 $\displaystyle \sum_{k=2}^\infty\frac1{k\ln k\ln(\ln k)\ln(\ln(\ln k))}$, and so on.

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

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

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

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

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

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

EDIT: Although, the above calculation seems to imply that if we let $\displaystyle B=\{p~prime,~p\equiv3\mod4\}$ then $\displaystyle \sum_{p\in B}\frac{1}{p}$ diverges.

$\displaystyle \sum_{p \equiv 1 \; \text{mod} \; 4} \frac{1}{p} $ diverges because one can show that given $\displaystyle (a,k)=1 $ and $\displaystyle A = \{$ prime $\displaystyle p \; | \; p \equiv a \; \text{mod} \; k \}$, $\displaystyle \lim_{s \to 1^{+}} \frac{\sum_{p \in A} \frac{1}{p^s}}{\sum_{p} \frac{1}{p^s}} = \frac{1}{\phi(k)} $.

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

Originally Posted by redsoxfan325

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

$\displaystyle \phi(k) $ is Euler's totient function. In your proof, What is $\displaystyle I $?

Originally Posted by chiph588@

$\displaystyle \phi(k) $ is Euler's totient function. In your proof, What is $\displaystyle I $?

If $\displaystyle p\equiv1\mod4$, then $\displaystyle p=m^2+n^2$ for some $\displaystyle m,n\in\mathbb{N}$. $\displaystyle I$ is the subset of $\displaystyle \mathbb{N}$ that contains all of those $\displaystyle m$ and $\displaystyle n$. It might be all of $\displaystyle \mathbb{N}$, but I wasn't sure, and I could think of a good way to write it, so I wrote $\displaystyle I$.

I'd say this equality is incorrect:
$\displaystyle \sum_{p \in A} \frac{1}{p} = \sum_{m,n\in I\subset\mathbb{N}}\frac{1}{m^2+n^2}$. This is because for a fixed $\displaystyle m $ (or $\displaystyle n $), the latter sum doesn't take into account how many times the $\displaystyle m $ (or $\displaystyle n $) appear in forming an arbitrary $\displaystyle p $.