Question

Use the ratio or root test to determine if the following series are convergent or divergent. If the ratio or the root test fails, indicate why.

[infinity]
Σ 1/(3n)!
n=1

Answer 1

Using the ratio test, the series 1/(3n)! is shown to be convergent, as the limit of the ratio of successive terms is zero, which is less than 1.

Step-by-step explanation:

The student's question involves determining whether the series ∑ 1/(3n)! from n=1 to infinity is convergent or divergent. To analyze this series, we can employ the ratio test, which compares the ratio of successive terms in the series.

Consider the general term a_n = 1/(3n)! and the next term a_(n+1) = 1/(3(n+1))!. Using the ratio test, we calculate:

limit of (a_(n+1)/a_n) as n approaches infinity.

This gives us:

limit of [(1/(3(n+1))!) / (1/(3n)!)] as n approaches infinity = limit of (3n)!/(3(n+1))! as n approaches infinity

= limit of (3n)!/(3n+3)! as n approaches infinity.

Rewriting (3n+3)! as (3n+3)(3n+2)(3n+1)(3n)!, we get:

limit of (3n)!/[(3n+3)(3n+2)(3n+1)(3n)!] as n approaches infinity

= limit of 1/[(3n+3)(3n+2)(3n+1)] as n approaches infinity

= 0.

Since the calculated limit is less than 1, the ratio test confirms that the series is convergent.