`sum_(n=0)^oo (-1)^n/((2n+1)!)` Determine the convergence or divergence of the series.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`


`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:








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`



The `L=0` satisfies ratio test condition: `Llt1`  since `0lt1` .

Thus, the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` is absolutely convergent.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team