Proving the Bessel function formula by expanding its generator function. I am trying to show that the Bessel functions $J_n(x)$  have the form
$$J_n(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!(n+k)!}\bigg( \frac{x}{2} \bigg)^{n+2k},$$ from its generator function
$$G(x,t)=\sum_{n=-\infty}^{\infty}t^nJ_n(x)=e^{\frac{x}{2}(t-\frac{1}{t})}.$$
Expanding G(x,t) in a power series with respect to its exponent we get
$$G(x,t)=\sum_{n=0}^{\infty}\frac{1}{m!} \bigg( t - \frac{1}{t}\bigg)^m\bigg( \frac{x}{2} \bigg)^{m}.$$
By the binomial expansion, we have
$$G(x,t)=\sum_{m=0}^{\infty}\sum_{k=0}^{m}\frac{(-1)^k}{k!(m-k)!} t^{m-2k}\bigg( \frac{x}{2} \bigg)^{m}.$$
By setting $n=m-2k$, the $a_{mk}$ factor of the sum becomes
$$\frac{(-1)^k}{k!(n+k)!} t^{n}\bigg( \frac{x}{2} \bigg)^{n+2k}.$$
So I seem to be in the correct way, yet I don't know how to change the $$\sum_{m=0}^{\infty}\sum_{k=0}^{m}$$ expressed in terms of $k$ and $n$. Any ideas? 

You want to identity the coefficients of $t^n$. Your formula includes
$t^{m-2k}$. So you want to set $m-2k=n$. Eliminating $m$ your formula
is the sum of
$$\frac{(-1)^k}{k!(n+k)!}t^n\left(\frac x2\right)^{n+2k}$$
over $n$ and $k$ with $0\le k\le n+2k$, that is $k\ge\max(0,-n)$.
So for $n\ge0$ you get
$$J_n(x)=\sum_{k=0}^\infty\frac{(-1)^k}{k!(n+k)!}\left(\frac x2\right)^{n+2k}.$$
