Use the ratio test to determine whether sigma^infinity+n = 12 n2^n/(n + 2)! conv

Question

Use the ratio test to determine whether sigma^infinity+n = 12 n2^n/(n + 2)! converges or diverges. Find the ratio of successive terms. Write your answer as a fully simplified fraction. For 11 Greaterthanorequalto 12, 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, lithe 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

PART A

L = lim (an+1/an) = (n+1)*2n+1*(n+2)!/(n+3)!n2n

lim = (n+1)*2/(n+3)n

PART B

L = lim (2* (1+1/n))/(n+3) = 0

PART C

L=0<1 hence series converges

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