Absolute Convergence and Divergence — Question 8

*Question 8 Consider $\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{n!}{(2n)!}$.

1. Simplify the ratio of consecutive absolute terms.

2. Apply the Ratio Test and classify convergence.

3. Explain the effect of the doubled factorial.

*Question 8 – Solution

For $b_n=n!/(2n)!$, $$\frac{b_{n+1}}{b_n}=\frac{(n+1)!}{(2n+2)!}\frac{(2n)!}{n!} =\frac{n+1}{(2n+2)(2n+1)}=\frac1{2(2n+1)}.$$

The ratio tends to $0<1$, so $\sum b_n$ converges. Hence the original series converges absolutely.

Each step in $(2n)!$ introduces two new denominator factors but only one new numerator factor. This drives the ratio rapidly to zero.