# How do you determine the convergence or divergence of #Sigma (-1)^(n+1)(1*3*5***(2n-1))/(1*4*7***(3n-2))# from #[1,oo)#?

##### 1 Answer

Jan 12, 2017

This series is (absolutely) convergent.

#### Explanation:

This series can be seen to be convergent on several grounds.

A rough evaluation is that the ratio between successive terms tends to

More formally, we proceed as follows:

Let:

#a_n = prod_(k=1)^n (2k-1)/(3k-2)#

Then:

#a_(n+1)/a_n = (2(n+1)-1)/(3(n+1)-2) = (2n+1)/(3n+1) <= 3/4#

when

So:

#sum_(n=1)^N a_n <= sum_(n=1)^N (3/4)^(n-1) = (1-(3/4)^N)/(1-3/4) = 4-4(3/4)^N#

Then:

#lim_(N->oo) (3/4)^N = 0#

So:

#sum_(n=1)^oo a_n <= 4# converges.

So our example series is absolutely convergent, and hence convergent.