Is there an analytic continuation at $z=R$? Assume $f(z)$ is analytic in $D_R(0)$ (in other words, the corresponding power series $\sum_{k=0}^{\infty}a_kz^k$ has the convergence radius equal to R). Suppose we make an analytic continuation to a point $z = r + i\ 0$ with $r < R$ (see the diagram below) and consider the power series representation for f(z) centered at $z = r$. If it happens that the corresponding power series has the convergence radius exactly $R − r$ (the blue circle), In this case, can we claim that no analytic continuation exists at the point $z = R$? Thanks

That is correct. An analytic function has at least one singularity on it's circle of convergence, in the sense that there is at least one point on it's circle of convergence in which the function cannot be analytically continued in an open ball around this point. You can find a discussion on this here.
For the blue circle, this point cannot be any other point than $z=R$. Indeed, for all the other points we can simply take an open ball around the point contained in the big red circle, and the series of $f$ around $0$ as our analytic continuation.
