Dirichlet series for square root of Riemann Zeta function

Question

Can we obtain Dirichlet series for the function $\sqrt{\zeta(s)}$? Is it possible via Euler product for $\zeta(s)$?

Answer 1

If there is a solution then it will be the product of the Dirichlet series of terms $(a_n)_{n\ge 1}$ with itself and thus have to verify the Dirichlet convolution $\;\displaystyle\sum_{d|n}\;a_d\cdot a_{n/d}=1$.

From this we obtain $\,a_1=1\,$ and $\ \ \displaystyle 2\cdot a_n=1-\sum_{d|n,d>1,d<n}\;a_d\cdot a_{n/d},\ \ n>1$ and thus :

$$\sqrt{\zeta(s)}=1+\frac 1{2\cdot 2^s}+\frac 1{2\cdot 3^s}+\frac 3{8\cdot 4^s}+\frac 1{2\cdot 5^s}+\frac 1{4\cdot 6^s}+\frac 1{2\cdot 7^s}+\frac 5{16\cdot 8^s}+\frac 3{8\cdot 9^s}+\frac 1{4\cdot 10^s}\\+\frac 1{2\cdot 11^s}+\frac 3{16\cdot 12^s}+\frac 1{2\cdot 13^s}+\frac 1{4\cdot 14^s}+\frac 1{4\cdot 15^s}+\frac {35}{128\cdot 16^s}+\frac 1{2\cdot 17^s}+\frac 3{16\cdot 18^s}+\cdots$$

Which is OEIS A046643 and A046644 for the numerators and denominators with more references.

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For fun let's add the expansion of the cube root : $$\sqrt[3]{\zeta(s)}=1+\frac 1{3\cdot 2^s}+\frac 1{3\cdot 3^s}+\frac 2{9\cdot 4^s}+\frac 1{3\cdot 5^s}+\frac 1{9\cdot 6^s}+\frac 1{3\cdot 7^s}+\frac {14}{81\cdot 8^s}+\frac 2{9\cdot 9^s}+\frac 1{9\cdot 10^s}+\frac 1{3\cdot 11^s}+\frac 2{27\cdot 12^s}+\frac 1{3\cdot 13^s}+\frac 1{9\cdot 14^s}+\frac 1{9\cdot 15^s}+\frac {35}{243\cdot 16^s}+\frac 1{3\cdot 17^s}+\frac 2{27\cdot 18^s}+\cdots$$ quartic root : $$\sqrt[4]{\zeta(s)}=1+\frac 1{4\cdot 2^s}+\frac 1{4\cdot 3^s}+\frac 5{32\cdot 4^s}+\frac 1{4\cdot 5^s}+\frac 1{16\cdot 6^s}+\frac 1{4\cdot 7^s}+\frac {15}{128\cdot 8^s}+\frac 5{32\cdot 9^s}\\+\frac 1{16\cdot 10^s}+\frac 1{4\cdot 11^s}+\frac 5{128\cdot 12^s}+\frac 1{4\cdot 13^s}+\frac 1{16\cdot 14^s}+\frac 1{16\cdot 15^s}+\frac{195}{2048\cdot 16^s}+\frac 1{4\cdot 17^s}+..$$ and quintic root : $$\sqrt[5]{\zeta(s)}=1+\frac 1{5\cdot 2^s}+\frac 1{5\cdot 3^s}+\frac 3{25\cdot 4^s}+\frac 1{5\cdot 5^s}+\frac 1{25\cdot 6^s}+\frac 1{5\cdot 7^s}+\frac {11}{125\cdot 8^s}+\frac 3{25\cdot 9^s}+\frac 1{25\cdot 10^s}+\frac 1{5\cdot 11^s}+\frac 3{125\cdot 12^s}+\frac 1{5\cdot 13^s}+\frac 1{25\cdot 14^s}+\frac 1{25\cdot 15^s}+\frac {44}{625\cdot 16^s}+\frac 1{5\cdot 17^s}+.$$ $$ $$ Interesting! In fact the coefficient of the $k$-term of the Dirichet series for $\,\sqrt[N]{\zeta(s)}\,$ appears to be :

• $\displaystyle\dfrac 1{N^d}\quad$ if $k$ is the product of $d$ distinct primes.
• $\displaystyle\dfrac {\prod_{i=1}^{m-1} N+\frac 1i}{m\,N^m}\quad$ if $k$ is the $m$-th power of a prime ($m>1$).
• a product of the previous ones for products of powers of distinct primes !?

Answer 2

An alternative way is to use Taylor series: $$ \sqrt {\zeta (s)} = \sqrt {1 + (\zeta (s) - 1)} = 1 + \frac{1}{2}(\zeta (s) - 1) - \frac{1}{8}(\zeta (s) - 1)^2 + \cdots . $$ Now just substitute the Dirichlet series of the zeta function into this expansion, re-expand it and collect the terms involving $\frac{1}{n^s}$.

Answer 3

$$F(s)=\prod_p (1+\sum_{k\ge 1} {-1/2\choose k}(-1)^kp^{-sk})=\sum_{n\ge 1} a_nn^{-s}$$ is the unique Dirichlet series such that $F(s)^2=\zeta(s)$ and $F(\infty)=1$. It converges for $\Re(s) > 1$. It has an awful branch point at $s=1$ making the asymptotic of $\sum_{n\le x} a_n$ complicated.