The series $\displaystyle\sum \dfrac{z^n}{n^2}$ converges for $\lvert z\rvert<1$ by the ratio test, meaning that the dilogarithm function $\text{Li}_2(z),$ which is equal to the series $\displaystyle\sum \dfrac{z^n}{n^2}$ when it converges, is certainly analytic on $\lvert z\rvert<1$.
Similarly, by the ratio test, the series diverges for $\lvert z\rvert>1$. And we know that for a meromorphic function, the radius of convergence is always the distance from the center to the nearest singularity(wikipedia). Hence we should conclude that this function has a pole somewhere on $\lvert z\rvert=1$?
On the other hand, at the point $z=1$ it converges to $\pi^2/6$ (see Basel problem), it must also converge at the point $z=-1$, and it is stated on this question by Ben that the series is convergent on the whole circle $\lvert z\rvert=1$ (this confused me and prompted the question), and on a comment to an answer to a related question, one is reminded by user 23rd that if a function is analytic on a closed disk, then it is analytic on an open disk of larger radius. So we should conclude that this series is convergent for some $z$ with $\lvert z\rvert>1$? For the whole complex plane? In fact, this Wolfram alpha page does claim that the function is analytic on all of $\mathbb{C}$ (if I'm reading it correctly; it's very terse).
Actually that second related question (singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$) already contains the answer to my question: the dilogarithm is analytic on $\mathbb{C}\setminus [1,\infty)$. But I can't understand how that answer is consistent with the other remarks, and the Wolfram page. How is this situation reconciled? Could I get an explanation that's a little more detailed than what's already there?
The alternative definition (easily proven equivalent to the one you give for $\lvert z \rvert < 1$) $$ \operatorname{Li}_2 (z) = \int_0^z \frac{\log{(1-t)}}{t} \, dt, $$ where the integral can be taken to be along the ray joining $0$ to $z$ for $z \notin (1,\infty) $, shows that the dilogarithm has a branch point at $z=1$ (since the logarithm inside the integral has one): you can see that no extension across the cut is possible by computing the derivative across the cut, for example.
As we see here, it is possible for a power series to converge on its whole circle of convergence (since the series is absolutely convergent for $\lvert z \rvert \leqslant 1$), but for no larger $\lvert z \rvert$; in this case, because the branch cut "gets in the way" of defining an analytic extension to a larger convex domain.