Use the ratio test to determine whether Sigma^infinity_n=22 n4^n/(n + 1)! conver

Question

Use the ratio test to determine whether Sigma^infinity_n=22 n4^n/(n + 1)! converges or diverges. Find the ratio of successive terms. Write your answer as a fully simplified fraction. For n greaterthanorequalto 22, lim_n rightarrow infinity a_n + 1/|a_n| = lim_n rightarrow infinity Evaluate the limit in the previous part. Enter infinity as infinity and -infinity as -infinity. If the limit does not exist, enter DNE. lim_n rightarrow infinity a_n + 1/|a_n| = By the ratio test, does the series converge, diverge, or is the test inconclusive?

Explanation / Answer

a)

an=n4n/(n+1)!

an+1=(n+1)4n+1/(n+2)!

ratio test for convergence:

limn->|an+1/an|

=limn->|[(n+1)4n+1/(n+2)!]/[n4n/(n+1)!]|

=limn->|4(n+1)/n(n+2)|

b)

limn->|4(n+1)/n(n+2)|

=limn->|4n(1+(1/n))/n2(1+(2/n))|

=limn->|4(1+(1/n))/n(1+(2/n))|

=|4(1+(0))/(1+(0))|

=0 <1

c) by ratio test the series converge

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