# Thread: convergence of a series

1. ## convergence of a series

determine convergence and show all work

inf
Sigma [3*5*7...(2n+1)] / [18^n(2n-1)n!]
n=1

I have been trying to do this problem for several days I have managed to figure out all the others but this one has me stumped.

Kevin

2. Hi

Let's show the convergence using the ratio test. (it's often useful when the terms of the series are defined by fractions or products)

$\displaystyle \frac{a_{n+1}}{a_n}=a_{n+1}\cdot \frac{1}{a_n}=\frac{{\color{red}3\cdot 5\ldots(2n+1)}\cdot(2n+3)}{18^{n+1}(2n+1)(n+1)!}\c dot \frac{18^n(2n-1)n!}{{\color{red}3\cdot5\ldots\cdot(2n+1)}}$

Simplifying what's in red and using $\displaystyle (n+1)!=(n+1)\cdot n!$ and $\displaystyle 18^{n+1}=18\cdot 18^n$ :

$\displaystyle \frac{a_{n+1}}{a_n}=\frac{2n+3}{18\cdot{\color{red }18^n}(2n+1)\cdot (n+1)\cdot {\color{blue}n!}}\cdot {\color{red}18^n}(2n-1){\color{blue}n!}$

$\displaystyle \frac{a_{n+1}}{a_n}=\frac{(2n+3)(2n-1)}{18(2n+1)(n+1)}$

The series converges if $\displaystyle \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| <1$.

Hope that helps.