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 2/3 as n->oo. So the series will eventually converge faster than a geometric series with ratio (say) 3/4, regardless of the alternating signs. In other words, this series is absolutely convergent.

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 n >= 1

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.