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We say that a series $\sum_{n=1}^\infty a_n$ and the corresponding power series $f(x)=\sum_{n=1}^\infty a_nx^n$ belong to the Ramanujan class $R=1$ if $g(x)=f(x)-f(x^2)$ is Abel summable at $x=1$ (here). Does the analytic continuation of the Dirichlet series $F(s)=\sum_{n=1}^\infty a_n n^{-s}$ have a simple pole at $s=0$?

Example 1. The harmonic series $$ 1+\frac{1}{2}+\frac{1}{3}+\cdots~, $$ corresponding to $f(x)=-\ln(1-x)$, $g(x)=\ln(1+x)$ ($|x|<1$) (here) and $F(s)=\zeta(s+1)=\eta(s+1)/(1-2^{-s})$.

Example 2. The series $1+1+0+1+0+0+0+1+\cdots$ corresponding to $f(x)=\sum_{k=0}^\infty x^{2^k}$, $g(x)=x$ and $F(s)=1/(1-2^{-s})$.

Example 3. For $g(x)=x-x^2+x^3-\cdots=x/(1+x)$ ($|x|<1$) corresponding to the Grandi series $1-1+1-\cdots$ Abel summable to $1/2$, we have $a_n=1-m_2(n)$, where $m_2(n)$ is the multiplicity of 2 (the 2-adic order) of $n$ and $F(s)=\eta(s)/(1-2^{-s})$.

In general, for a power series $g(x)=\sum_{n=1}^\infty b_n x^n$ Abel summable at $x=1$, the coefficients $a_n$ are determined by the recursive relations $a_{2n-1}=b_{2n-1}$ and $a_{2n}=b_{2n}+a_n$. Is the analytic continuation of $(1-2^{-s})F(s)$ holomorphic at $s=0$?

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No. The analytic continuation of $F(s)$ can be holomorphic at $s=0$ if $\lim_{x\to 1^-}g(x)=0$.

Example 1. $f(x)=x$, $g(x)=x-x^2$, $F(s)=1$ for all $s$.

Example 2. $g(x)=x-x^3$, $f(x)=\sum_{k=0}^\infty x^{2^k}-x^{3\times2^k}$, $F(s)=(1-3^{-s})/(1-2^{-s})$, with $F(0)=\ln3/\ln2$ and $F'(0)=-\ln3\times\ln(3/2)/2\ln2$.

If $\lim_{x\to 1^-}g(x)\neq 0$, the analytic continuation of $F(s)$ has a simple pole at $s=0$, since the recursive relation between the coefficients of $f(x)$ and $g(x)$ implies $F(s)=G(s)/(1-2^{-s})$, where $G(s)=\sum_{n=1}^\infty b_n n^{-s}$ (here and here).

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