 Solution

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.