# `sum_(n=0)^oo (2x)^n` Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) Recall the Root test determines the limit as:

`lim_(n-gtoo) root(n)(|a_n|)= L`

or

`lim_(n-gtoo) |(a_n)|^(1/n)= L`

a) `Llt1` then the series is absolutely convergent

b) `Lgt1` then the series is divergent.

c) `L=1` or does not exist  then the test is inconclusive. The series may be divergent, conditionally convergent, or absolutely convergent.

For the given series `sum_(n=0)^oo (2x)^n` , we have `a_n = (2x)^n` .

Applying the Root test, we set-up the limit as:

`lim_(n-gtoo) |((2x)^n )^(1/n)| =lim_(n-gtoo) |(2x)^(n*1/n)|`

`=lim_(n-gtoo) |(2x)^(n/n)|`

`=lim_(n-gtoo) |(2x)^1|`

`=lim_(n-gtoo) |(2x)|`

` =|2x|`

Applying  `Llt1` as the condition for an absolutely convergent series, we let `L=|2x|` and set-up the interval of convergence as:

`|2x|lt1`

`-1 lt2xlt1`

Divide each part by 2:

`(-1)/2 lt(2x)/2lt1/2`

`-1/2ltxlt1/2`

The series may converges when `L =1` or `|2x|=1` . To check on this, we test for convergence at the endpoints: `x=-1/2` and `x=1/2` by using geometric series test.

The convergence test for the geometric series `sum_(n=0)^oo a*r^n`  follows the conditions:

a) If `|r|lt1`  or `-1 ltrlt 1` then the geometric series converges to `a/(1-r)` .

b) If `|r|gt=1` then the geometric series diverges.

When we let `x=-1/2` on `sum_(n=0)^oo (2*(1/2))^n ` , we get a series:

` sum_(n=0)^oo 1*(-2/2)^n =sum_(n=0)^oo 1*(-1)^n`

It shows that `r=-1` and `|r|= |-1|=1` which satisfies |r|>=1. The series diverges at the left endpoint.

When we let `x=1/2` on `sum_(n=0)^oo (2*1/2)^n` , we get a series:

`sum_(n=0)^oo 1*(2/2)^n =sum_(n=0)^oo 1*(1)^n`

It shows that `r=1` and `|r|= |-1|=1` which satisfies `|r|gt=1` . The series diverges at the right endpoint.

Conclusion:

The interval of convergence of the power series `sum_(n=0)^oo (2x)^n ` is `-1/2ltxlt1/2` .

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