Question

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


`sum (x - 2)^(2) \ (ni.{2}^(n))`
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Answer 1

By the ratio test, the series converges if

`\displaystyle\lim_(n\to\infty)\left|(((x-2)^(n+1))/((n+1)!2^(n+1)))/(((x-2)^n)/(n!2^n))\right|=|x-2|\lim_(n\to\infty)(n!2^n)/((n+1)!2^(n+1))`

`=\displaystyle\fracx-22\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)`.