Is there any relationship between a Dirichlet series and the same series with the sequence "shifted" by one term?

Question

Suppose $$ F(s) := \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} $$ is a Dirichlet series for the sequence $a_{1}, a_{2}, \ldots\in\mathbb{C}$. Then let $$ G(s) := \sum_{n=1}^{\infty} \frac{a_{n}}{(n+1)^{s}} $$ be the same series but with the sequence "shifted" by one index. Is there any known relationship between $F(s)$ and $G(s)$?

My question seems to be a subcase of this question, but I think my question should stand on its own, because the levels of generality here are different enough to warrant possibly different answers.

I'm okay with multiple suggestions.

Answer 1

There was an excellent suggestion made by user reuns that was later refined by Sangchul Lee.

Proposition. Let $\sigma_a$ be the abscissa of absolute convergence of $F$. Then for each $s$ with $\Re(s) > \sigma$, $$ \sum_{k=0}^{\infty} \binom{-s}{k} [F(s+k) - a_1] = \sum_{n=2}^{\infty} \frac{a_n}{(n+1)^s}, $$ where the convergence of the series is absolute. Moreover, $$ \lim_{\varepsilon \to 0^+} \sum_{k=0}^{\infty} \binom{-s}{k} e^{-\varepsilon k} F(s+k) = \sum_{n=1}^{\infty} \frac{a_n}{(n+1)^s} $$ (In other words, the proposed identity in OP holds in Abel summation sense.)

This is taken from this post here. For the sake of completeness, I will post their proof here as well.

To prove the first claim, note that

$$ \binom{-s}{k} = \frac{(-1)^k}{\Gamma(s)} \frac{\Gamma(s+k)}{\Gamma(k+1)} \sim \frac{(-1)^k}{\Gamma(s)} k^{s-1} \qquad \text{as } k \to \infty. $$

So, there exists a constant $C = C(s) \in (0, \infty)$ such that $\left|\binom{-s}{k}\right|\leq C (k + 1)^{\Re(s)-1}$ for all $k \geq 0$. Then by this bound and the Tonelli's theorem together, we get

\begin{align*} \sum_{k=0}^{\infty} \sum_{n=2}^{\infty} \left| \binom{-s}{k} \frac{a_n}{n^{s+k}} \right| &\leq C \sum_{k=0}^{\infty} \sum_{n=2}^{\infty} \frac{(k + 1)^{\Re(s)-1} |a_n|}{n^{\Re(s)+k}} \\ &\leq C \biggl( \sum_{k=0}^{\infty} \frac{(k + 1)^{\Re(s)-1}}{2^k} \biggr)\biggl( \sum_{n=2}^{\infty} \frac{|a_n|}{n^{\Re(s)}} \biggr) < \infty. \end{align*}

So, by the Fubini's theorem,

\begin{align*} \sum_{k=0}^{\infty} \binom{-s}{k} [F(s+k) - a_1] &= \sum_{n=2}^{\infty} \sum_{k=0}^{\infty} \binom{-s}{k} \frac{a_n}{n^{s+k}} \\ &= \sum_{n=2}^{\infty} \frac{a_n}{n^s(1 + \frac{1}{n})^s} = \sum_{n=2}^{\infty} \frac{a_n}{(n+1)^s}. \end{align*}

The second claim then follows from the Abel's theorem for power series and

$$ \lim_{\varepsilon \to 0^+} \sum_{k=0}^{\infty} \binom{-s}{k} e^{-\varepsilon k} a_1 = \lim_{\varepsilon \to 0^+} \frac{a_1}{(1 + e^{-\varepsilon})^s} = \frac{a_1}{2^s}. $$