Convergence and geometric tails for integer-order Bessel series

For $z\mathrm{\ne }0$, every term is nonzero, and cancellation of the factorials gives $\mathrm{\mid }{a}_{k+1}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{a}_k\mathrm{\mid }=\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}[4(k+1)(k+n+1)]$. This ratio decreases to $0$. At $z=0$, the series have only the possible order-zero constant term, so every tail after $k=0$ is zero without using a quotient.

For $z\mathrm{\ne }0$ and $k\ge K+1$ the ratios are at most $q$, so ${\sum }_{k=K+1}^{\mathrm{\infty }}\mathrm{\mid }{a}_k\mathrm{\mid }\le \mathrm{\mid }{a}_{K+1}\mathrm{\mid }{\sum }_{j=0}^{\mathrm{\infty }}{q}^j=\mathrm{\mid }{a}_{K+1}\mathrm{\mid }\mathrm{/}(1-q)$. At $z=0$ the same tail bound holds because the tail is zero. On each disk $\mathrm{\mid }z\mathrm{\mid }\le R$, put $q_R=R^2\mathrm{/}[4(K+2)(K+n+2)]$. For sufficiently large $K$, $q_R<1$ and the nonzero-argument tail is bounded uniformly by ${\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R}\mathrm{\mid }{a}_{K+1}(z)\mathrm{\mid }\mathrm{/}(1-q_R)$; at $z=0$ it is zero. Thus the series converge uniformly on every disk.

For derivative order $d\in \{1,2\}$, put $m=2k+n$. When $m\ge d$, the absolute value of the formal derivative term on $\mathrm{\mid }z\mathrm{\mid }\le R$ is at most $b_k=m(m-1)\cdots (m-d+1)R^{m-d}\mathrm{/}(2^{m}k!(k+n)!)$. For $R>0$ the ratio $b_{k+1}\mathrm{/}{b}_k$ equals $R^2\mathrm{/}[4(k+1)(k+n+1)]$ times a factor tending to $1$, so it tends to $0$. The ratio test makes $\sum b_k$ finite; the Weierstrass test then gives uniform convergence of each formal derivative series on the disk. The finitely many terms with $m<d$ have zero derivative, and a disk of radius $R=0$ is immediate.