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`sum_(n=0)^oo (2n)!(x/3)^n` Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.)

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For the power series `sum_(n=0)^oo (2n)!(x/3)^n` , 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.

The given power series `sum_(n=0)^oo (2n)!(x/3)^n` has:

`a_n =(2n)!(x/3)^n`

Then,

`1/a_n=1/((2n)!)(3/x)^n`

      ` =1/((2n)!)(3^n/x^n)`

      ` =3^n/((2n)!x^n)`

`a_(n+1) =(2(n+1))!(x/3)^(n+1)`

            ` = (2n+2)!x^(n+1)/3^(n+1)`

            `= (2n+2)(2n+1)((2n)!) x^n*x/(3^n*3)`

             `=((2n+2)(2n+1)((2n)!) * x^n*x)/(3^n*3)`

Applying the Ratio test on the power series, we set-up the limit as:

`lim_(n-gtoo) |((2n+2)(2n+1)((2n)!) * x^n*x)/(3^n*3)3^n/((2n)!x^n)|`

Cancel out common factors: `x^n` , `(2n)!` , and `3^n` .

`lim_(n-gtoo) |((2n+2)(2n+1)*x)/3|`

Evaluate the limit.

`lim_(n-gtoo) |((2n+2)(2n+1)*x)/3| = |x/3|lim_(n-gtoo) |(2n+2)(2n+1)|`

                                            `= |x/3|* oo`

                                           ` = oo `

The limit value `L= oo` satisfies` Lgt 1` for all `x` . 

Therefore,  the power series `sum_(n=0)^oo (2n)!(x/3)^n` diverges for all x based from the Ratio test criteria:  `Lgt1 ` then the series diverges.

There is no interval for convergence.

Note: The radius of convergence is `0` . The `x=0` satisfy the convergence at points where `(2n)!(x/3)^n=0` .

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