= Solution
Let $a_n=(1+p/n)^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,
$$
\ln|a_n|^{1/n}=n\ln\left(1+\frac pn\right)=p-\frac{p^2}{2n}+O(n^{-2})\longrightarrow p.
$$
The <Cauchy-Hadamard theorem> therefore yields
$$
\boxed{R=e^{-p}\qquad\text{for every real }p.}
$$
For example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.
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