How do you apply the ratio test to determine if #sum_(n=1)^oo (e^n(n!))/n^n# is convergent or divergent?

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Answer 1 · 2017-05-02T21:38:21

Please see the explanation.

Given: sum_(n=1)^oo (e^n(n!))/n^n

Now for the ratio test :

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

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

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

#L = lim_(nrarroo)|( (e^(n+1)((n+1)!))/(n+1)^(n+1))/((e^n(n!))/n^n)|#

#L = lim_(nrarroo)|(e^(n+1)((n+1)!))/(n+1)^(n+1)(n^n)/(e^n(n!))|#

#L = lim_(nrarroo)|((e)e^n(n!(n+1)))/(n+1)^(n+1)(n^n)/(e^n(n!))|#

#L = lim_(nrarroo)|((e)cancel(e^n)(cancel(n!)cancel((n+1))))/(n+1)^(ncancel(+1))(n^n)/(cancel(e^n)(cancel(n!)))|#

#L = lim_(nrarroo)|e(n/(n+1))^n|#

#L = 1#

The ratio test is inconclusive.

However, if you do the limit test:

#L = lim_(nrarroo)(e^n(n!))/n^n = oo#

This shows that the sum diverges.