for the following power series, find the interval of absolute convergence `sum_(n=1)^(oo) ((2x)^n)/(n!)`

1 Answer | Add Yours

mlehuzzah's profile pic

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

 

The series converges absolutely if

 

`lim _(n->oo) (|a_(n+1)|)/(|a_n|) <1`

`|a_n| = |((2x)^n)/(n!)|`

`|a_(n+1)| = |((2x)^(n+1))/((n+1)!)|`

So:
`(|a_(n+1)|)/(|a_n|)`

`= |((2x)^(n+1))/((n+1)!) (n!)/((2x)^n) |`

`= |(2x)/(n+1)|`

`=(2)/(n+1) |x|`

Now, pick any x. x=-3, x=17, x=10000000. It doesn't matter. No matter what, once you pick that x, then it becomes just a constant. Now, let `n -> oo`

As `n->oo`, `1/(n+1) -> 0` , so:

`lim _(n->oo) (2)/(n+1) |x|`

`= 2|x| lim _(n->oo) (1)/(n+1) = 0`

Since `0<1` this converges. But that was true of ANY x

So for any x, the power series converges. Thus the radius of convergence is `oo` and the interval of convergence is `(-oo, oo)`

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question