`sum_(n=0)^oo (-1)^n/((2n+1)!)` Determine the convergence or divergence of the series.
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To determine the convergence or divergence of the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` , we may apply ratio test.
In Ratio test, we determine the limit as:
`lim_(n-gtoo)|a_(n+1)/a_n| = L`
or
`lim_(n-gtoo)|a_(n+1)*1/a_n| = L`
Then ,we follow the conditions:
a) `L lt1` then the series converges absolutely.
b) `Lgt1` then the series diverges.
c)` L=1` or does not exist then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.
For the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` , we have:
`a_n=(-1)^n/((2n+1)!)`
Then,
`1/a_n=((2n+1)!)/(-1)^n`
`a_(n+1)=(-1)^(n+1)/((2(n+1)+1)!)`
`=(-1)^(n+1)/((2n+2+1)!)`
`=(-1)^(n+1)/((2n+3)!)`
`=((-1)^n*(-1))/((2n+3)(2n+2)((2n+1)!))`
Applying the Ratio test on the power series, we set-up the limit as:
`lim_(n-gtoo) |((-1)^n*(-1))/((2n+3)(2n+2)((2n+1)!)) *((2n+1)!)/(-1)^n|`
Cancel out common factors: `(-1)^n` and `(2n+1)!` .
`lim_(n-gtoo) |(-1)/((2n+3)(2n+2))|`
Evaluate the limit.
`lim_(n-gtoo) |(-1)/((2n+3)(2n+2))| =|-1| lim_(n-gtoo) |1/((2n+3)(2n+2))|`
`=1* 1/oo`
`=1*0`
`=0`
The `L=0` satisfies ratio test condition: `Llt1` since `0lt1` .
Thus, the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` is absolutely convergent.
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