Find the interval of converge to this series? Sum when n=1 and goes to infinity (x-2) ^n/ (n! .2^)
[tex] sum (x - 2)^{2} \ (ni.{2}^{n}) [/tex]

Respuesta :

By the ratio test, the series converges if

[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(x-2)^{n+1}}{(n+1)!2^{n+1}}}{\frac{(x-2)^n}{n!2^n}}\right|=|x-2|\lim_{n\to\infty}\frac{n!2^n}{(n+1)!2^{n+1}}[/tex]

[tex]=\displaystyle\frac{|x-2|}2\lim_{n\to\infty}\frac1{n+1}[/tex]

is less than 1. The limit itself is 0 < 1, so the series converges everywhere, i.e. on the entire real line [tex](-\infty,\infty)[/tex].