# 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)?

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 $\frac{2}{3}$ as $n \to \infty$. So the series will eventually converge faster than a geometric series with ratio (say) $\frac{3}{4}$, regardless of the alternating signs. In other words, this series is absolutely convergent.

More formally, we proceed as follows:

Let:

${a}_{n} = {\prod}_{k = 1}^{n} \frac{2 k - 1}{3 k - 2}$

Then:

${a}_{n + 1} / {a}_{n} = \frac{2 \left(n + 1\right) - 1}{3 \left(n + 1\right) - 2} = \frac{2 n + 1}{3 n + 1} \le \frac{3}{4}$

when $n \ge 1$

So:

${\sum}_{n = 1}^{N} {a}_{n} \le {\sum}_{n = 1}^{N} {\left(\frac{3}{4}\right)}^{n - 1} = \frac{1 - {\left(\frac{3}{4}\right)}^{N}}{1 - \frac{3}{4}} = 4 - 4 {\left(\frac{3}{4}\right)}^{N}$

Then:

${\lim}_{N \to \infty} {\left(\frac{3}{4}\right)}^{N} = 0$

So:

${\sum}_{n = 1}^{\infty} {a}_{n} \le 4$ converges.

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