A squarefree function?

Number Theory Level 5

$\large \sum_{n=2}^\infty \dfrac{\ln (\varsigma(n))}{n^2}$

Let $\varsigma(n)$ be the greatest squarefree divisor of $n$ . For example, $\varsigma(8) = 2, \varsigma(10) = 10, \varsigma(12) = 6$ . Evaluate the summation above.

Give your answer to 3 decimal places.

You may use a calculator such as WolframAlpha for the final calculation.

The answer is 0.8111.

1 solution

Julian Poon
Jan 2, 2016

Convergence proof:

$\sum _{ n>1 }{ \frac { \ln { \varsigma (n) } }{ { n }^{ 2 } } } <\sum _{ n>1 }{ \frac { \ln { n } }{ { n }^{ 2 } } } <\sum _{ n>1 }{ \frac { \sqrt { n } }{ { n }^{ 2 } } } =\zeta (1.5)$

Of which $\zeta(1.5)$ is known to converge.

First thing to notice is that

$\ln\varsigma (n)=\sum _{ p|n }{ \ln { p } } =|\mu |\Lambda *1(n)$

Where $\mu$ is the Mobius function , $\Lambda$ is the Von Mangoldt function and $f*g$ is the Dirichlet convolution.

So the sum that we are finding is the Dirichlet series :

$DG(|\mu |\Lambda *1;2)$

Using the multiplication of Dirichlet series, $DG(f;s)DG(g;s)=DG(f*g;s)$

$DG(|\mu |\Lambda *1;2) = DG(|\mu |\Lambda ;2) \times \frac{\pi^{2}}{6}$

Calculating numerically,

$DG(|\mu |\Lambda ;2)=\sum _{p \text{ is prime}}\frac{\ln p}{p^2}=-0.493091...$

Hence the sum is equal to $-0.493091...\cdot \frac{\pi ^2}{6}=0.81110...$

Moderator note:

Nice application of the Dirichlet convolution to calculate such sums. One could also see it directly, by asking "When does the term $\ln p$ appear in the numerator?" and realize that we collect terms of the form

$\frac{1}{p^2 } + \frac{1}{(2p)^2} + \frac{1}{(3p)^2} + \ldots$

It is slightly unfortunate that there isn't a nice way to deal with $\sum \frac{\ln p } { p^2}$ , which is essentially the equivalent problem.

As an aside, my favorite proof of the convergence of $\sum \frac{ \ln n } { n^2}$ is using the integral test. (Can you evaluate $\int \frac{\ln n } { n^2 }$ ? It shows that the value is bounded above by 1.

In fact, this works for $\frac{\ln n } { n^a }$ , where $a > 1$ .

Calvin Lin Staff - 5 years, 5 months ago

i reached the part where this was $\dfrac{\pi^2}{6}\sum_{p=prime} \dfrac{\ln(p)}{p^2}$ then it became $\dfrac{\pi^2}{6}\sum_{n=1}^\infty \dfrac{\Lambda(n)}{n^2}$ by its connection to zeta function we have it equal to $-\dfrac{\pi^2}{6}*\dfrac{\zeta'(2)}{\zeta(2)}\approx .937$ can you help me @Julian Poon

Aareyan Manzoor - 5 years, 5 months ago

Why would turn into $\Lambda(n)$ since $\Lambda(p^a)=p$ instead of $0$ ?

Julian Poon - 5 years, 5 months ago

oh! thats where stuff went wrong! damn i knew i should have used dirtchlet... nice solution btw.

Aareyan Manzoor - 5 years, 5 months ago

Nice! Exactly as intended. In fact, the closed form is $-\zeta(2)\mathcal{P}'(2)$ where $\mathcal{P}$ is the prime zeta function, which makes it easier to calculate.

Jake Lai - 5 years, 5 months ago

Is there a way of calculating P'(x) with Mr. wolf? Because I just used

$-\frac{\zeta'\left(2\right)}{\zeta\left(2\right)}+\sum _{k=2}^{\infty}\frac{\zeta'\left(2k\right)}{\zeta\left(2k\right)}$

EDIT: Oh wait, just type d/dx PrimeZetaP(x), x=2

Julian Poon - 5 years, 5 months ago

1
2
3

>>> import pyprimes, math
>>> sum(math.log(p) / p**2 for p in pyprimes.primes_below(100000)) * math.pi**2 / 6
# 0.8110859425287458

And yet here we have

$\displaystyle\begin{aligned} 0.81108\ldots &\approx \frac{\pi^2}{6} \sum_{p\text{ prime}, 2 \le p \le 100000} \frac{\ln p}{p^2} \\ &< \frac{\pi^2}{6} \sum_{p\text{ prime}} \frac{\ln p}{p^2} \\ &= \sum_{p\text{ prime}} \sum_{k=1}^\infty \frac{\ln p}{(pk)^2} \\ &= \sum_{n=2}^\infty \sum_{p\text{ prime}, p|n} \frac{\ln p}{n^2} \\ &= \sum_{n=2}^\infty \frac{\ln \varsigma(n)}{n^2} \\ &\approx 0.80576\ldots \end{aligned}$

Considering you multiplied with $\frac{\pi^2}{6}$ like me, I just assume you made a mistake somewhere in computing the one that gave you $0.489\ldots$ .

Ivan Koswara - 5 years, 5 months ago

Wait, I'm confused, isn't the sum $\sum \frac{\ln \varsigma(n)}{n^2}$ rather than $\sum \frac{\varsigma(n)}{n^2}$ ? Also, the true value of $\sum_p \frac{\ln p}{p^2}$ is -0.493091 if I'm not wrong, so Julian made an underestimation.

Jake Lai - 5 years, 5 months ago

Sorry, that last line was a typo. It is indeed $\sum \frac{\ln \varsigma(n)}{n^2}$ ; I just typed incorrectly.

Ivan Koswara - 5 years, 5 months ago

I think Wolfram Alpha made an understatement in calculating

Julian Poon - 5 years, 5 months ago

i just realised i would have gotten this right if i "numerically" calculated the last part instead of using zeta function... damn.

Aareyan Manzoor - 5 years, 5 months ago

A squarefree function?

The answer is 0.8111.

1 solution

Moderator note:

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