Prove that $ B_{n}(x)=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}$ converges uniformly. Let $n\geq 0,$ be a fixed. The Bessel function of order $n$ is the function defined by
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{align}
I want to prove that $B_{n}(x)$ converges uniformly on any closed interval $[a,b]\subseteq \Bbb{R}.$
MY WORK
Let $[a,b]\subseteq \Bbb{R}$ be arbitrary, $x\in [a,b]$ and $n\in \Bbb{N}$ be fixed. Then,
\begin{align} \left| \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}\right|\leq  \left( \dfrac{\left|  x\right|} {2}\right)^{n+2k}\leq   \left(\dfrac{c}{2}\right)^{n+2k}.\end{align}
where $c=\max\{|a|,|b|\}.$ Now,
\begin{align} \sum^{\infty}_{k=0} \left(\dfrac{c}{2}\right)^{n+2k} =\left(\dfrac{c}{2}\right)^{n}\sum^{\infty}_{k=0} \left(\dfrac{c}{2}\right)^{2k} =\left(\dfrac{c}{2}\right)^{n} \left(\dfrac{4}{4-c^2}\right)<\infty,\;\;\text{where}\;\;c<2.\end{align}
Hence,
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{align}
converges uniformly on $[a,b]\subseteq \Bbb{R}$ with $c<2.$
QUESTION:
The question asks for a proof for any closed interval of $ \Bbb{R}.$ With what I have, it only works when $c<2.$ What happens when $c\geq 2$? Or I'm I missing something in the proof?
 A: No, your proof surely does not work for all intervals $[a,b]$ of $\mathbb R$. It only works for those intervals $[a,b]$ such that $\lvert a\rvert,\lvert b\rvert<2$.
A: Happy New Year! Credits to Jakobian 
\begin{align} \left| \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k}\right|\leq  \dfrac{1} {k!}\left( \dfrac{\left|  x\right|} {2}\right)^{n+2k}\leq   \dfrac{1} {k!}\left(\dfrac{c}{2}\right)^{n+2k}.\end{align}
Since,
\begin{align} \lim\limits_{k\to\infty}\left|\dfrac{k!} {(k+1)!}\left(\dfrac{c}{2}\right)^{n+2k+2} \left(\dfrac{2}{c}\right)^{n+2k} \right|&=\lim\limits_{k\to\infty}\left|\dfrac{1} {(k+1)}\left(\dfrac{c}{2}\right)^{2}  \right|\\&=\dfrac{c^2}{4}\lim\limits_{k\to\infty}\dfrac{1} {(k+1)} \\&=0<1\end{align}
Thus, $\sum^{\infty}_{k=0}    \dfrac{1} {k!}\left(\dfrac{c}{2}\right)^{n+2k}$ converges by D'Alembert's Ratio test and so,
\begin{align} B_{n}(x):=\sum^{\infty}_{k=0} \dfrac{(-1)^k}{k!(n+k)!} \left(\dfrac{x}{2}\right)^{n+2k} .\end{align}
converges uniformly on any $[a,b]\subseteq \Bbb{R}$
