summation (3^n)( n!)/(n^n) does the above series converge or diverge when n goes to infinity?

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summation (3^n)( n!)/(n^n) does the above series converge or diverge when n goes to infinity?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Use Ratio test
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i know that but i m getting it to be convergent when it is a divergent series

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ya it is divergent the limit is 3e^(-1) which is larger than 1
how did u get e?
Ratio test lim |a_(n+1)/a_n| = lim 3[n/(n+1)]^n
o yea... so n/n+1)^n = e right
sorry 1/e
yup, the limit is 1/e
Do you know how to get that limit?

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