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