Solution: Because ρ=n→+∞lim[a? n+ 1？ /a？ n？ ]=n→+∞lim[ 1/(2n+2))！ ]/[ 1/(2n)！ ]

=n→+∞lim[(2n)！ /(2n+2)！ ]=n→+∞lim[(2n)！ /(2n)！ (2n+ 1)(2n+2)]

= n →+∞lim[ 1/(2n+ 1)(2n+2)]= 0

So the convergence radius R= 1/ρ=∞, and the convergence interval is (-∞, +∞).