How do you apply the ratio test to determine if $Sigma (n!)/n^n$ from $n=[1,oo)$ is convergent to divergent?

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Answer 1 · 2017-02-23T12:56:30

The series:

$sum_(n=1)^oo (n!)/n^n$

is convergent.

Evaluate the ratio:

$abs(a_(n+1)/a_n) = ( ((n+1)!)/(n+1)^(n+1))/ ((n!)/n^n) = n^n/(n+1)^(n+1) ((n+1)!)/(n!) = 1/(n+1)(n/(n+1))^n (n+1)=(n/(n+1))^n$

We can write this expression differently in order to find the limit:

$(n/(n+1))^n = (1/((n+1)/n))^n = 1/(1+1/n)^n$

So:

$lim_(n->oo) abs(a_(n+1)/a_n) = 1/(1+1/n)^n = 1/e <1$

which proves the series to be convergent.

How do you apply the ratio test to determine if Sigma (n!)/n^nSigma (n!)/n^n from n=[1,oo)n=[1,oo) is convergent to divergent?