= Solution
For $y''+P(x)y'+Q(x)y=0$, a singular point $x_0$ is a <regular singular point> if $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are analytic there. The <Bessel differential equation> has $xP=1$ and $x^2Q=x^2-\nu^2$, so zero is regular singular.
Use the <Frobenius method> with $y=x^s\sum_{n\geq0}a_nx^n$, $a_0\ne0$. Matching coefficients gives $(s^2-\nu^2)a_0=0$ and $[(s+n)^2-\nu^2]a_n+a_{n-2}=0$, with missing negative-index coefficients zero. The regular root is $s=\nu$. Then $a_1=0$, all odd coefficients vanish, and the <Frobenius series for Bessel functions> has recurrence
$$
\boxed{a_{2m}=-\frac{a_{2m-2}}{4m(m+\nu)},\qquad m\geq1.}
$$
For noninteger $\nu$, the root $s=-\nu$ supplies an independent even-power series $y_2=x^{-\nu}\sum_{m\geq0}b_{2m}x^{2m}$ with $b_{2m}=-b_{2m-2}/[4m(m-\nu)]$ and $b_0\ne0$. None of these denominators vanishes. With standard normalization this is $J_{-\nu}(x)$, independent of the <Bessel function of the first kind> $J_\nu(x)$. The real powers here are understood for $x>0$.
For integer $\nu\geq0$ and the stated normalization $a_0=1/(2^\nu\nu!)$, iteration gives
$$
\boxed{a_{2m}=\frac{(-1)^m}{2^{2m+\nu}m!(m+\nu)!},\qquad J_\nu(x)=\sum_{m=0}^\infty\frac{(-1)^m(x/2)^{2m+\nu}}{m!(m+\nu)!}.}
$$
The ratio of consecutive term magnitudes is $|x|^2/[4(m+1)(m+\nu+1)]\to0$. Thus the series converges absolutely for every finite complex $x$, defining an entire function for integer $\nu$. At integer order, $J_{-\nu}=(-1)^\nu J_\nu$; an independent solution is instead the <Bessel function of the second kind>.
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