`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.)

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`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)` .

Approved by eNotes Editorial Team