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#