Question

Find the interval of converge to this series? Sum when n=1 and goes to infinity (x-2) ^n/ (n! .2^)

Answer 1

By the ratio test, the series converges if

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

$=\displaystyle\frac{|x-2|}2\lim_{n\to\infty}\frac1{n+1}$

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