`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.)

Expert Answers

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`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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