Issue proving a dubious identity involving Dirichlet series Let $a_{1}, a_{2}, \ldots\in\mathbb{C}$ be a sequence, and let
$$ F(s) := \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} \qquad\text{ and }\qquad G(s) := \sum_{n=1}^{\infty} \frac{a_{n}}{(n+1)^{s}}. $$
Someone suggested to me that the two Dirichlet series are related by
$$ G(s) \stackrel{?}{=} \sum_{k=0}^{\infty}\binom{-s}{k}F(s+k). $$
I am trying to prove this, but I ran into trouble.
Claim. Let
$$ F(s) = \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} \qquad\text{ and }\qquad G(s) = \sum_{n=1}^{\infty} \frac{a_{n}}{(n+1)^{s}}. $$
If $s > \max(0, \sigma_{a})$ where $\sigma_{a}$ is the abscissa of absolute convergence of $F$, then we have
$$ G(s) = \sum_{k=0}^{\infty}\binom{-s}{k}F(s+k). $$
Attempted Proof. We have
\begin{align*}
\sum_{k=0}^{\infty} \binom{-s}{k}F(s+k) &= \sum_{k=0}^{\infty} \binom{-s}{k} \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s+k}} \\[1.4ex]
&= \sum_{k=0}^{\infty} \sum_{n=1}^{\infty} \binom{-s}{k} \frac{a_{n}}{n^{s+k}} \\[1.4ex]
&= \sum_{\color{red}{k=0}}^{\infty} \sum_{\color{red}{n=2}}^{\infty} \binom{-s}{k} \frac{a_{n}}{n^{s+k}} + \color{red}{\sum_{\color{red}{k=0}}^{\infty} \binom{-s}{k} a_{1}} \\[1.4ex]
&= \sum_{\color{red}{n=2}}^{\infty} \sum_{\color{red}{k=0}}^{\infty} \binom{-s}{k} \frac{a_{n}}{n^{s+k}} + \color{red}{\sum_{k=0}^{\infty} \binom{-s}{k} a_{1}} \\[1.4ex]
&= \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \binom{-s}{k} \frac{a_{n}}{n^{s+k}} \\[1.4ex]
&= \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} \sum_{k=0}^{\infty} \binom{-s}{k} \frac{1}{n^{k}} \\[1.4ex]
&= \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} \left(1 + \frac{1}{n} \right)^{-s} \\[1.4ex]
&= \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} \left( \frac{n+1}{n}\right)^{-s} \\[1.4ex]
&= \sum_{n=1}^{\infty} \frac{a_{n}}{(n+1)^{s}} \\[1.4ex]
&= G(s).
\end{align*}
First, there is the issue of interchanging summation signs between lines 3 and 4. I believe this can be taken care of by a version of some dominated convergence theorem, and I think there are no problems there.
However, I've taken out some sums in those lines and highlighted them in red. Since we're assuming $s > \max(0, \sigma_{a})$, those sums seem to be divergent. If $a_{1}\ne 0$, this seems to be a serious problem.
My questions are,

*

*Where did I go wrong with this proof? What is the proper way to proceed rigorously?

*Is the claim even true?

 A: As in OP's setting, let $a_1, a_2, \ldots \in \mathbb{C}$ and suppose $F(s)$ is defined by
$$ F(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s}. $$
Then the following claim is true:

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.)

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