sum_(n=0)^oo x^(5n)/(n!) 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 x^(5n)/(n!)
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, the value of L will be:
L= lim_(n->oo) |(x^(5(n+1))/((n+1)!))/(x^(5n)/(n!))|
L= lim_(n->oo)|x^(5(n+1))/((n+1)!) * (n!)/x^(5n)|
L= lim_(n->oo) | x^(5n+5)/((n+1)n!) * (n!)/x^(5n)|
L= lim_(n->oo) |x^5/(n+1)|
L = x^5* lim_(n->oo) |1/(n+1)|
L = x^5 * 0
L=0
Since the value of L is less than 1, the given series converges for all values of x.
Therefore, the interval of convergence is (-oo, oo) .