Im trying to prove that $$e^{\frac{x}{2}(t-t^{-1})}=\sum_{n=-\infty}^{\infty}t^nJ_{n}(x)$$ Where $$J_n(x)=\sum_{s=0}^{\infty}\frac{(-1)^s}{s!(s+n)!}\left(\frac{x}{2}\right)^{2s+n}$$ Is the Bessel function of order n.
Using the power series of the exponential and expanding $t-t^{-1}$ via the binomial theorem I have that $$\sum_{m=0}^{\infty}\frac{1}{m!}\left(\frac{x}{2}\right)^m\sum_{s=0}^{m}\frac{m!}{(m-s)!s!}t^{m-s}(-t^{-1})^s$$
$$\sum_{m=0}^{\infty}\left(\frac{x}{2}\right)^m\sum_{s=0}^{m}\frac{(-1)^s}{(m-s)!s!}t^{m-2s}$$
Then setting $r=m-s$
$$e^{\frac{x}{2}(t-t^{-1})}=\sum_{m=0}^{\infty}\sum_{r+s=m}\frac{(-1)^s}{r!s!}\left(\frac{x}{2}\right)^{r+s}t^{r-s}$$
This is where I start having trouble. Specifically my issue is with the summation indices.
My attempt so far is that for a general summation: $$\sum_{m=0}^{\infty}\sum_{r+s=m}a_{r,s}=(a_{0,0})+(a_{0,1}+a_{1,0})+(a_{0,2}+a_{1,1}+a_{2,0})+...$$$$=(a_{0,0}+a_{0,1}+a_{0,2}+...)+(a_{1,0}+a_{1,1}+a_{1,2}+...)+...=\sum_{r=0}^{\infty}\sum_{s=0}^{\infty}a_{r,s}$$
So that
$$e^{\frac{x}{2}(t-t^{-1})}=\sum_{r=0}^{\infty}\sum_{s=0}^{\infty}\frac{(-1)^s}{r!s!}\left(\frac{x}{2}\right)^{r+s}t^{r-s}$$
I'm not sure about my argument on this, so if theres a better/more rigorous way to do this I'd love to hear it. From this point I set $n=r-s$ to get
$$e^{\frac{x}{2}(t-t^{-1})}=\sum_{n=-s}^{\infty}\sum_{s=0}^{\infty}\frac{(-1)^s}{(n+s)!s!}\left(\frac{x}{2}\right)^{n+2s}t^{n}=\sum_{n=-s}^{\infty}t^nJ_n(x)$$
This id so close to the correct result, but no matter what I do I can't seem to get the correct limits of $\pm \infty$ on the final sum. Any help with this (and with double summation limit interchange in general), would be much appreciated.
I’m not sure about your counting error, or where you are making it so perhaps seeing the calculation might help you find it. Therefore consider $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{m=0}^{\infty }{\frac{{{x}^{m}}}{{{2}^{m}}m!}{{\left( t-\frac{1}{t} \right)}^{m}}}=\sum\limits_{m=0}^{\infty }{\frac{{{\left( \tfrac{1}{2}x \right)}^{m}}}{m!}\sum\limits_{s=0}^{m}{\left( \begin{matrix} m \\ s \\ \end{matrix} \right){{t}^{m-s}}{{\left( -t \right)}^{-s}}}}$$
Therefore $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{m=0}^{\infty }{{{\left( \tfrac{1}{2}x \right)}^{m}}{{t}^{m}}\sum\limits_{s=0}^{m}{\frac{{{\left( -1 \right)}^{s}}}{s!\left( m-s \right)!}{{t}^{-2s}}}}$$ Swapping the order of summation we have $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{s=0}^{\infty }{\frac{{{\left( -1 \right)}^{s}}}{s!\left( m-s \right)!}{{t}^{-2s}}}\sum\limits_{m=s}^{\infty }{{{\left( \tfrac{1}{2}x \right)}^{m}}{{t}^{m}}}$$ or $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{s=0}^{\infty }{\frac{{{\left( -1 \right)}^{s}}{{\left( \tfrac{1}{2}x \right)}^{s}}}{s!}{{t}^{-s}}}\sum\limits_{m=0}^{\infty }{\frac{{{\left( \tfrac{1}{2}x \right)}^{m}}}{m!}{{t}^{m}}}$$ Perhaps this is where you are going awry? If so what we have used here is the following result $$\sum\limits_{k=0}^{\infty }{\sum\limits_{j=0}^{k}{{{a}_{k,j}}}}=\sum\limits_{j=0}^{\infty }{\sum\limits_{k=j}^{\infty }{{{a}_{k,j}}}}$$ To see this result you can either do as you have done, by evaluating several terms and then regrouping them, or you can visualise the double series as a two dimensional array of numbers where order of summation is going first along rows/columns and then columns/rows etc (much the same as changing order of integration in a double integral). A third way is to use Iverson brackets. There is a fourth way, and that is to observe that the s-summation can actually be over $0\le s<\infty $ since the binomial coefficients become zero when s>m. Doing that then removes any issues with order. Anyway…continuing…
We want powers of t, so consider the coefficient against a general ${{t}^{n}}$ term for $n\ge 0$. We need the $m=n+s$ term from the second series and so the coefficient is therefore $${{J}_{n}}\left( x \right)=\sum\limits_{s=0}^{\infty }{\frac{{{\left( -1 \right)}^{s}}{{\left( \tfrac{1}{2}x \right)}^{2s+n}}}{s!\left( n+s \right)!}}$$ where we choose to label these coefficients as ${{J}_{n}}\left( x \right)$. Consider now the coefficient against a general ${{t}^{-n}}$ term where $n>0$. We need the $s=n+m$ term from the first series. Hence the coefficients will be $$\sum\limits_{m=0}^{\infty }{\frac{{{\left( -1 \right)}^{n+m}}{{\left( \tfrac{1}{2}x \right)}^{2m+s}}}{\left( m+n \right)!m!}}={{J}_{-n}}\left( x \right)={{\left( -1 \right)}^{n}}{{J}_{n}}\left( x \right)$$ We now have $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{m=0}^{\infty }{{{J}_{m}}\left( x \right){{t}^{m}}}+\sum\limits_{m=1}^{\infty }{{{J}_{-m}}\left( x \right){{t}^{-m}}}=\sum\limits_{m=-\infty }^{\infty }{{{J}_{m}}\left( x \right){{t}^{m}}}$$ where $${{J}_{n}}\left( x \right)=\sum\limits_{s=0}^{\infty }{\frac{{{\left( -1 \right)}^{s}}{{\left( \tfrac{1}{2}x \right)}^{2s+n}}}{s!\left( n+s \right)!}}$$ and $${{J}_{-n}}\left( x \right)={{\left( -1 \right)}^{n}}{{J}_{n}}\left( x \right)$$ As per my comment, if you don't specifically need to use binomial coefficients then I think the typical manner in which to show this result is to a) break the generating function into two exponentials and then use their power series expansion and apply the same reasoning as above or b) residues. Since a) leads to the exact same arguments as those above, for completeness, let’s examine b). Consider the Laurent expansion of the function $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{m=-\infty }^{\infty }{{{J}_{m}}\left( x \right){{t}^{m}}}$$ where we label the coefficients on t in the expansion as ${{J}_{m}}\left( x \right)$. We have therefore $${{J}_{m}}\left( x \right)=\frac{1}{2\pi i}\int\limits_{C}^{{}}{\frac{1}{{{s}^{m+1}}}{{e}^{\tfrac{1}{2}x\left( s-\frac{1}{s} \right)}}ds}$$ where the contour circles the origin once. We have $${{J}_{m}}\left( x \right)=\frac{1}{2\pi i}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{x}^{k}}}{{{2}^{k}}k!}}\int\limits_{C}^{{}}{\frac{1}{{{s}^{m+k+1}}}{{e}^{\tfrac{1}{2}xs}}ds}$$ Using the substitution $t=\tfrac{1}{2}xs$ $${{J}_{m}}\left( x \right)=\frac{1}{2\pi i}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}x \right)}^{2k+m}}}{k!}}\int\limits_{C}^{{}}{\frac{1}{{{t}^{m+k+1}}}{{e}^{t}}dt}$$ Now this has residues of $1/\left( m+k \right)!$ when m+k is a positive integer or zero, and zero residue when m+k is a negative integer. If m, therefore, is a positive integer, or zero, we have then $${{J}_{m}}\left( x \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}x \right)}^{2k+m}}}{k!\left( m+k \right)!}}$$ If, however, m is a negative integer, say -n, then we have $${{J}_{-n}}\left( x \right)=\frac{1}{2\pi i}\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}x \right)}^{2k-n}}}{k!}}\int\limits_{C}^{{}}{\frac{1}{{{t}^{k-n+1}}}{{e}^{t}}dt}$$ This has residues $1/\left( k-n \right)!$ but only for $k\ge n$. So $${{J}_{-n}}\left( x \right)=\sum\limits_{k=n}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}x \right)}^{2k-n}}}{k!\left( k-n \right)!}}=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k+n}}{{\left( \tfrac{1}{2}x \right)}^{2k+n}}}{\left( k+n \right)!k!}=}{{\left( -1 \right)}^{n}}{{J}_{n}}\left( x \right)$$ therefore $${{e}^{\tfrac{1}{2}x\left( t-\frac{1}{t} \right)}}=\sum\limits_{m=-\infty }^{\infty }{{{J}_{m}}\left( x \right){{t}^{m}}}$$ where $${{J}_{m}}\left( x \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( \tfrac{1}{2}x \right)}^{2k+m}}}{k!\left( m+k \right)!}},\,\,\,{{J}_{-m}}\left( x \right)={{\left( -1 \right)}^{m}}{{J}_{m}}\left( x \right)$$