$F(s)=\sum_{n=1}^\infty \frac{f(n)}{n^s}$ and $G(s)=\sum_{n=1}^\infty \frac{g(n)}{n^s}$ s.t $F(s_k)=G(s_k)$ show that $f(n)=g(n)$

Question

Let $F(s)=\sum_{n=1}^\infty \frac{f(n)}{n^s}$ and $G(s)=\sum_{n=1}^\infty \frac{g(n)}{n^s}$ be two Dirichlet series which are absolutely convergent for $\Re(s)>a$ for some $a\in \Bbb R$. If there is a sequence $\{s_k\}\subset \Bbb C$ with $\Re(s_k)\to \infty$ as $k \to \infty$ s.t $F(s_k)=G(s_k)$ for all $k$, show that $f(n)=g(n)$ for all $n$.

I was first thinking to use induction or proceed by contradiction. But why will $f(1)=g(1)$. I am not getting it anyway. Please help. Also share if you have some other idea.

Answer 1

Let $b$ be any number bigger than $a$ and $c_k= \mathcal{R}(s_k)$
So $$\lim c_k = +\infty$$

If $f(p)=g(p)$ for all $ 1 \le p \le m$ with $m$ being any positive integer, for any $k$ big enough, we have:

$$\begin{align} \left| (f(m+1)-g(m+1)) \right| &\stackrel{(1)}{=} \left| (m+1)^{s_k}\sum_{n \ge m+2} \frac{f(n)-g(n)}{n^{s_k}}\right|\le \sum_{n \ge m+2} |f(n)-g(n)| ( \frac{m+1}{n})^{c_k} \\ &\stackrel{(2)}{\le} \sum_{n \ge m+2} |f(n)-g(n)| ( \frac{m+1}{n})^{b}( \frac{m+1}{m+2})^{c_k-b} \\ &= ( \frac{m+1}{m+2})^{c_k-b} (m+1)^b \underbrace{\sum_{n \ge m+2} \frac{|f(n)-g(n)|}{n^b}}_{<\infty \quad (3)} \xrightarrow[]{k \rightarrow+\infty} 0 \end{align}$$ where:

• In (1), we use the fact that $F(s_k)=G(s_k)$
• In (2), we use the fact that $c_k-b>0$ when $k$ is big enough
• In (3), we use the absolute convergence

So $f(m+1)=g(m+1)$ if $f(p)=g(p)$ for all $ 1 \le p \le m$.

The base case $m=1$,in fact, can be shown in the same manner.