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

maeveoneill

Junior Member
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Sep 24, 2005
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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.
 
For a given 'n', consider the sequence of successive multipliers.

After n!(2)n135...(2n1)\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: B  =  A(n+1)(2)2n+1  =  2n+22n+1  >  A\displaystyle B\;=\;A*\frac{(n+1)*(2)}{2n+1}\;=\;\frac{2n+2}{2n+1}\;>\;A

And the next: B(n+2)(2)2n+3  =  2n+42n+3  >  B\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.
 
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