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

*print*Print*list*Cite

### 1 Answer

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.