You probably know that
[tex]\sin x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]
Then
[tex]\mathrm{sinc}\,x=\displaystyle\frac1x\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}[/tex]
when [tex]x\neq0[/tex], and 1 when [tex]x=0[/tex].
By the ratio test, the series converges if the following limit is less than 1:
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+2}}{(2n+3)!}}{\frac{(-1)^nx^{2n}}{(2n+1)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)!}{(2n+3)!}[/tex]
The limit is 0, so the series converges for all [tex]x[/tex].
