= Solution
The absolute ratio of successive terms is $|z|\,n^2/(n+1)^2\to|z|$. The <ratio test> gives convergence for $|z|<1$ and divergence for $|z|>1$, so the <radius of convergence> is $\boxed{R=1}$. On the boundary, $|n^{-2}z^n|=n^{-2}$ and
$$
\sum_{n=2}^N\frac1{n^2}\le\sum_{n=2}^N\frac1{n(n-1)}=1-\frac1N.
$$
The positive partial sums are bounded, and thus converge. By the <comparison test for series>, the original <power series> \b[converges absolutely at every point of $|z|=1$].
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