Let $\mathbb H= \{z \in \mathbb C\,:\, \textrm{Re}\,(z)\geq 0\}$ and for $j=1,2$ suppose that $F^{(j)}:\mathbb H\to \mathbb H$ is defined via
$$ F^{(j)}(z) = \sum\_{k=1}^{\infty} \frac{a\_k^{(j)}}{z+\lambda^{(j)}\_k},$$
where $|a\_{k}^{(j)}|\leq k^{-2}$ and $0<\lambda^{(j)}\_1<\lambda^{(j)}\_2<...$ with $\lim\_{k\to \infty} \lambda^{(j)}\_k=\infty$. Suppose that
$$ F^{(1)}(n)=F^{(2)}(n)\quad \text{for all $n \in \mathbb N$}.$$
Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?



Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?
The answer is positive. Indeed, $f=F^{(1)}-F^{(2)}$ is a bounded analytic function in the right half-plane (this follows from your conditions $|a\_k|\leq k^{-2}$ and $\lambda\_k\to\infty$). But a bounded analytic function cannot
be zero at positive integers, unless it is identically equal to zero.
This follows from the [Blaschke condition](https://en.wikipedia.org/wiki/Blaschke_product) on zeros of bounded analytic functions.


Your assumption that $F$ maps the right half-plane into itself is redundant, and the assumption that $\lambda\_k>0$ can be very much relaxed; all you need is that your functions are bounded in the right half-plane.


