Use Ratio Test to find the convergence of the following series?

sum_(n=1)^oo ((n+3)!)/ ((3!)(n!)(3^n)

Apr 1, 2018

The series:

sum_(n=0)^oo (( n+3)!) / ( ( 3!)(n!)(3^n))

is convergent.

Explanation:

Evaluate the ratio:

a_(n+1)/a_n = abs ( ( ( (n+1+3)!)/((3!)((n+1)!)3^(n+1)) ) / ( (( n+3)!) / ( ( 3!)(n!)(3^n)))

a_(n+1)/a_n = ((n+4)!)/((n+3)!) (3!)/(3!) (n!)/((n+1)!) 3^n/3^(n+1)

${a}_{n + 1} / {a}_{n} = \frac{1}{3} \frac{n + 4}{n + 1}$

Hence:

${\lim}_{n \to \infty} {a}_{n + 1} / {a}_{n} = \frac{1}{3}$

and the series is convergent.