[sp]That argument leaves me feeling a bit queasy, but I believe it gives the correct answer. Here's another approach.
Form the power series $$y = \frac{1}{3}x^3 +\frac{2}{3\cdot5}x^5 +\frac{2\cdot4}{3\cdot5\cdot7}x^7 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9}x^9 + \ldots.$$
Differentiate: $$y' = x^2 + \frac{2}{3}x^4 +\frac{2\cdot4}{3\cdot5}x^6 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^8 + \ldots.$$
Then $$\begin{aligned} \frac{y' - x^2}{x^3} &= \frac13(2x) + \frac{2}{3\cdot5}(4x^3) + \frac{2\cdot4}{3\cdot5\cdot7}(6x^5) + \ldots \\ &= \frac{d}{dx}\left(\frac yx\right) \\ &= \frac{xy'-y}{x^2}, \\ y' - x^2 &= x^2y'- xy, \\ y'(1-x^2) + xy &= x^2.\end{aligned}$$
Solve that differential equation by using an integrating factor $(1-x^2)^{-3/2}$, to get $y'(1-x^2)^{-1/2} + x(1-x^2)^{-3/2}y = x^2(1-x^2)^{-3/2}.$ That integrates to give $$\frac y{(1-x^2)^{1/2}} = \int\frac{x^2}{(1-x^2)^{3/2}}dx = \int \tan^2\theta\, d\theta = \int( \sec^2\theta - 1)\,d\theta = \tan\theta - \theta = \frac x{\sqrt{1-x^2}} - \arcsin x,$$ using the substitution $x = \sin\theta$ and choosing the constant of integration so that $y=0$ when $x=0$.
So $y = x - \sqrt{1-x^2}\arcsin x.$ When $x=1$, $y=1$. But when $x=1$ the power series for $y$ becomes $$\frac{1}{3} +\frac{2}{3\cdot5} +\frac{2\cdot4}{3\cdot5\cdot7} + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9} + \ldots.$$ This shows that $$1 = \frac{1}{3} +\frac{2}{3\cdot5} +\frac{2\cdot4}{3\cdot5\cdot7} + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9} + \ldots$$, and subtracting $\dfrac13$ from both sides gives the required answer.
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