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`sum_(n=1)^oo` n! e^-4n       does this converge or diverge

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danielwhitlock | eNotes Newbie

Posted March 25, 2013 at 6:58 PM via web

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`sum_(n=1)^oo` n! e^-4n       does this converge or diverge

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pramodpandey | College Teacher | Valedictorian

Posted March 27, 2013 at 6:58 AM (Answer #1)

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Let    `a_n=n!e^(-4n)`

`a_(n+1)=(n+1)!e^(-4(n+1))`

`a_(n+1)/a_n={(n+1)!e^(-4(n+1))}/{n!e^(-4n)}`

`=(n+1)e^(-4)`

`lim_(n->oo){a_(n+1)/a_n}=lim_(n->oo){(n+1)/e^4}`

`=oo`

Therefore by d'Alembert ,ratio test given series diverges.

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