Question

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)`?

Answer 1

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.