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.)

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 .