[SPLIT] test n!(-2)^n / 1.3.5...(2n-1) for convergence

[maeveoneill]

What test would you use to find out if n!(-2)^n / 1.3.5...(2n-1) is convergent??

I thought the ratio test, but it gives 1, which is inconclusive.

 

[tkhunny]

For a given 'n', consider the sequence of successive multipliers.

After \(\displaystyle \frac{n!*(-2)^{n}}{1*3*5*...*(2n-1)}\), call that 'A',

I'll ignore the sign on the 2.

The next term is: \(\displaystyle B\;=\;A*\frac{(n+1)*(2)}{2n+1}\;=\;\frac{2n+2}{2n+1}\;>\;A\)

And the next: \(\displaystyle B*\frac{(n+2)*(2)}{2n+3}\;=\;\frac{2n+4}{2n+3}\;>\;B\)

These terms are not decreasing, although in the limit the ratio is 1.