Convergence of a series given in non-closed form

[Entertainment Unit]

Homework Statement

Determine whether the given series is absolutely convergent, conditionally convergent, or divergent.

##\frac{1}{3} + \frac{1 \cdot 4}{3 \cdot 5} + \frac{1 \cdot 4 \cdot 7}{3 \cdot 5 \cdot 7} + \frac{1 \cdot 4 \cdot 7 \cdot 10}{3 \cdot 5 \cdot 7 \cdot 9} + \ldots + \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)} + \ldots##

Homework Equations

None that I'm aware.

The Attempt at a Solution

Before I can apply any of the convergence tests, I need a closed-form expression.

##a_n = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)}##

##= \frac{2 \cdot 1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)}{2 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)}##

##= \frac{2 \cdot 1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)(4 \cdot 6 \cdot \ldots \cdot 2n)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot \ldots \cdot 2n(2n + 1)}##

##= \frac{2 \cdot 1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)2(2 \cdot 3 \cdot \ldots \cdot n)}{(2n + 1)!}##

##= \frac{2 \cdot 1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)2n!}{(2n + 1)!}##

##= \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)4n!}{(2n + 1)!}##

and then I'm not sure where to go.

 

[andrewkirk]

You don't need to do much calculation in this. Just work out whether the terms are ultimately increasing or decreasing. If they are not decreasing then the series must be divergent, since they are all positive. To prove that, just find the smallest term, and use the fact that all terms are at least as great as that.

 

[Entertainment Unit]

Thanks, here's what wound up with:

It is given that ##a_n = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)}##

##\implies a_{n + 1} = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)(3n + 1)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)(2n + 3)}##

Suppose ##a_n \lt a_{n + 1}##

##\frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)} \lt \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)(3n + 1)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)(2n + 3)}##

##1 \lt \frac{3n+1}{2n + 3}##

##n \gt 2##

which means our supposition that ##a_n \lt a_{n+1}## is correct for ##n > N = 2 \implies \lim_{n\to\infty} a_n = \infty \implies## the given series is divergent by the Test For Divergence.