Does #sum_(n=1)^oo 3^n/(n!)# converge ?

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#sum_(n=1)^oo 3^n/(n!)#

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Answer 1 · 2017-06-04T16:01:29

Yes. By the ratio test

#lim_(nrarroo)|A_(n+1)/A_n|#

#A_n = 3^n/(n!)#

#A_(n+1) = (3/(n+1))3^n/(n!)#

#lim_(nrarroo)|3/(n+1)| = 0#

I asked WolframAlpha; it confirmed my answer and gave me a value:

#sum_(n=1)^oo 3^n/(n!) = e^3-1#