How do you find the interval of convergence $Sigma (x-5)^n/(n!)$ from $n=[0,oo)$ ?

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Answer 1 · 2017-03-05T20:19:34

The series:

$sum_(n=0)^oo (x-5)^n/(n!)$

is convergent for any $x in RR$

Apply the ratio test:

$lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( ( (x-5)^(n+1)/((n+1)!))/ ((x-5)^n/(n!)))$

$lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( (x-5)^(n+1)/(x-5)^n) (n!)/((n+1)!)$

$lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs (x-5) /(n+1) = 0$

So the series is convergent for any $x in RR$

How do you find the interval of convergence Sigma (x-5)^n/(n!)Sigma (x-5)^n/(n!) from n=[0,oo)n=[0,oo)?