2
$\begingroup$

Following the example to compute $\zeta (s)\sum_{n=1}^{\infty}\frac{\phi(n)}{n^s}=\zeta (s-1)$, converges absolutely if $\sigma>2$, where $\phi(n)$ is the Euler's totient function and $s=\sigma + it$ is a complex variable, I am trying compute the Dirichlet series for the radical of an integer, too called its squarefree kernel, caution I assume this multiplicative function: $rad(1)=1$ and for $n=\prod_{i=1}^{k}p_{i}^{e_{i}}$, the factorization in primes of $n>1$ then $rad(n)=\prod_{i=1}^{k}p_{i}$, it is the product of distinct primes which divide $n$.

My attempt: The Dirichlet series $\sum_{n=1}^{\infty}\frac{rad(n)}{n^s}$, could be difficult to obtain because I don't known compute in a closed form certain relatede amounts.

1) I compute for $\sigma>2$, if my computations aren't wrong $$\sum_{n=1}^{\infty}\frac{rad(n)}{n^s}=1+\frac{1}{\zeta (s)}\sum_{m=1}^{\omega(n)}\frac{D_{m}(n)}{n^{s}}$$ where $\omega(n)=k$ is the number of distinct primes that divide $n$, and $D_{1}(n)=\sum_{i=1}^{k}e_{i}p_{i}$, $D_{2}(n)=\sum_{i\neq j}e_{i}e_{j}p_{i}p_{j}$, $D_{3}(n)=\sum_{i\neq j, j\neq l, i\neq l}e_{i}e_{j}e_{l}p_{i}p_{j}p_{l}$,...,$D_{\omega(n)}(n)=\prod_{i=1}^{k}e_{i}p_{i}$, with absolute convergence, with $\sigma>2$ for all Dirichlet series $\sum D_{m}(n)/n^{s}$. The problem is sum, or bound in terms of another Dirichlet series these, to continue with computations. 2) After I've computed for $\sigma>2$, $\sum_{n=1}^{\infty}\frac{rad(n)}{n^s}$ as a Euler's product, since $rad$ and $\phi$ are mulplicative functions, and $\zeta(s)$ don't vanishing I write (really I believe that it could be wrong because I don't know if a character $\chi$ can be assumed equal to $-\phi$, it is if has sense $\chi(n)=-\phi(\widehat{n})$), for $\sigma>2$ and Euler's product of $R(s)=\sum_{n=1}^{\infty}\frac{rad(n)}{n^s}$

$$R(s)=\prod_{\text{p,prime}}\frac{p^{s}-1+p}{p^{s}-1}$$ that $$\zeta(s)\cdot L(s,\chi)=R(s)$$

really adjusting factors from related Euler's products of realated Dirichlet series of $\mu(n)$, Mobius function and $\chi(n)$, the previous related Dirichlet character that satisfies if it is possible, or has sense $\chi=-\phi$

Thanks in advice. If you known some reference for these computations please add the reference or hints to continue in the way. I am reading [1], and my only goal is study and understand this theory, and that you, or I, edit this post with the rights computations. Really I don't known if previous relation has sense or what is its significate, in such case.

Question.

A) Can you continue (with sums or bounds that showing something more about $\sum_{n=1}^{\infty}rad(n)/n^s$), improve or refute my computations in 1).

B) Can you refute $\zeta(s)\cdot L(s,\chi)=R(s)$ in 2)? Has sense define the Dirichlet character as I said?

References: [1] Tom M. Apostal, Introduction to Analytic Number Theory, Springer (1976).

$\endgroup$
  • $\begingroup$ What is $\sigma$??? It never appear in some formula. $\endgroup$ – Masacroso Jul 17 '15 at 14:22
  • 1
    $\begingroup$ @Masacroso $\sigma = \operatorname{Re} s$. $\endgroup$ – Daniel Fischer Jul 17 '15 at 14:22
  • $\begingroup$ I am sorry Masacroso, thanks Fischer, $\endgroup$ – user243301 Jul 17 '15 at 14:23
  • $\begingroup$ Thanks for vote up my question. $\endgroup$ – user243301 Nov 11 '15 at 11:42
4
$\begingroup$

You haven't exploited the main feature of the multiplicativity of $\text{rad}(n)$: the Euler's product.

$$\sum_{n\geq 1}\frac{\text{rad}(n)}{n^s}=\prod_{p}\left(1+p\cdot p^{-s}+p\cdot p^{-2s}+\ldots\right)=\prod_p\left(1+\frac{p}{p^s-1}\right).$$

By approximating $\frac{p}{p^s-1}$ with $\frac{1}{p^{s-1}}$, we have:

$$\sum_{n\geq 1}\frac{\text{rad}(n)}{n^s}\approx \prod_{p}\left(1+\frac{1}{p^{s-1}}\right)=\frac{\zeta(s-1)}{\zeta(2s-2)}.$$

$\endgroup$
  • $\begingroup$ You have reason, I tried sum in the product, your answer is accept. Really I don't know if my computations fail, if you see some bad computation, please say me in future days. My only goal is learn in this ME. $\endgroup$ – user243301 Jul 17 '15 at 15:19

Your Answer

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