A complex valued function $F,$ defined on an open set $E$ in the plane $\mathbb R^{2}$, is said to be real-analytic in $E$ if to every point $(s_{0}, t_{0})$ in there corresponds an expansion with complex coefficients $$F(s, t)= \sum_{n,m=0}^{\infty} a_{nm}(s-s_{0})^{m} (t-t_{0})^{n},$$ which converges absolutely for all $(s,t)$ in some neighbourhood of $(s_{0}, t_{0}).$
If $F$ is defined in the whole plane $\mathbb R^{2}$ by a series $$F(s, t)= \sum_{n,m=0}^{\infty} a_{nm}s^{m} t^{n},$$ which converges absolutely for every $(s,t),$ then we call $F$ real-entire.
My Question is: How to show
$$f(s,t) = \frac{1}{(1+s^{2}) (1+t^{2})}, (s,t \in \mathbb R)$$
is real- analytic in the whole plane $\mathbb R^{2}$ but not real-entire.
Thanks,
If $(s,t)\mapsto f(s,t)$ is real-entire in $\mathbb R^2$, then plugging $t=0$ into the power series we get a real-entire function $s\mapsto f(s,0)$.
In this case $f(s,0)=\dfrac{1}{1+s^2} $, which is not real-entire. The radius of convergence of the power series at $0$ is $1$.
To show your function is real-analytic, use the fact that $f(x) = 1/(1+x^2)$ is real-analytic. Which can itself be proved in various ways, e.g. by general theorem that $f/g$ is real-analytic on the set $\{g\ne 0\}$ provided $f,g$ are real analytic.