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

sum_(n=0)^oo (3x)^n/((2n)!)
To determine the interval of convergence, use Ratio Test.  The formula in Ratio Test is:
L = lim_(n->oo) |a_(n+1)/a_n|
If L <1, the series is absolutely convergent. 
If L>1, the series is divergent.
And if L = 1, the test is inconclusive. The series may converge or diverge.
Applying the formula above, the value of L will be:
L = lim_(n->oo) |(((3x)^(n+1))/((2(n+1))!))/(((3x)^n)/((2n)!))|
L= lim_(n->oo) | ((3x)^(n+1))/((2(n+1))!)* ((2n)!)/((3x)^n)|
L= lim_(n->oo) | ((3x)^(n+1))/((2n+2)!)* ((2n)!)/((3x)^n)|
L= lim_(n->oo) | ((3x)^(n+1))/((2n+2)(2n+1)(2n)!)* ((2n)!)/((3x)^n)|
L= lim_(n-gtoo) | (3x)/((2n+2)(2n+1))|
L=3x lim_(n->oo) |1/((2n+2)(2n+1))|
L=3x*0
L=0
Since the value of L is less than 1, the given series converges for any values of x.
Therefore, the interval of convergence is (-oo, oo) .