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

Question

Isabella Ackerman · Accepted Answer

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.

Adam Adkins · Answer

undefined To apply the ratio test to determine the convergence or divergence of the series $ \sum_{n=1}^\infty \frac{e^n(n!)}{n^n} $, you would calculate the limit of the ratio of consecutive terms as $ n $ approaches infinity:

$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

where $ a_n = \frac{e^n(n!)}{n^n} $.

1. First, compute $ a_{n+1} $ and $ a_n $.
2. Then, simplify the expression $ \frac{a_{n+1}}{a_n} $.
3. Finally, evaluate the limit $ L $.

If $ L < 1 $, the series converges. If $ L > 1 $, the series diverges. If $ L = 1 $, the test is inconclusive, and another test or method may be needed.