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`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...

`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)` .

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