Question

Consider the series \sum_{n=1}^{\infty} \frac{33^n (n!)^3}{(3n)!} \\ Let a_n = \frac{33^n (n!)^3}{(3n)!}. When applying the ratio test to this series: \\ (a) The ratio \frac{a_{n+1}}{a_n} simplifies to: \frac{a_{n+1}}{a_n} = \boxed{} Give your answer fully simplified. \\ (b) Evaluate the limit: \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \boxed{} \\ (c) Complete the conclusion: \\ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \boxed{<} , so the series \sum_{n=1}^{\infty} \frac{33^n (n!)^3}{(3n)!} \boxed{} by the \boxed{}

          Consider the series \sum_{n=1}^{\infty} \frac{33^n (n!)^3}{(3n)!} \\ Let a_n = \frac{33^n (n!)^3}{(3n)!}. When applying the ratio test to this series: \\ (a) The ratio \frac{a_{n+1}}{a_n} simplifies to: \frac{a_{n+1}}{a_n} = \boxed{} Give your answer fully simplified. \\ (b) Evaluate the limit: \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \boxed{} \\ (c) Complete the conclusion: \\ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \boxed{<} , so the series \sum_{n=1}^{\infty} \frac{33^n (n!)^3}{(3n)!} \boxed{} by the \boxed{}

Consider the series ∑n=1^∞ (33^n (n!)^3)/((3n)!)
Let an = (33^n (n!)^3)/((3n)!). When applying the ratio test to this series:
(a) The ratio (an+1)/(an) simplifies to: (an+1)/(an) = Give your answer fully simplified.
(b) Evaluate the limit: limn →∞ (an+1)/(an) =
(c) Complete the conclusion:
limn →∞ (an+1)/(an) < , so the series ∑n=1^∞ (33^n (n!)^3)/((3n)!) by the

Added by Christopher Y.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

08/18/2023

Step 1

To find the ratio an+1/an, we need to find the value of the term an+1 and divide it by the value of the term an. The term an+1 is given by (3(n+1))! / 33((n+1)!)³. The term an is given by (3n)! / 33(n!)³. So, the ratio an+1/an is: ((3(n+1))! / 33((n+1)!)³) / Show more…

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Texts: Consider the series (3n)! 33(n!)³. Let an be the term of the series. When applying the ratio test to this series: a) The ratio an+1/an. Give your answer fully simplified. b) Evaluate the limit: lim n→∞ an. (c) Complete the conclusion: The series converges, so the series is convergent.