# `sum_(n=0)^oo (x/4)^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) |(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 (x/4)^n` , we have `a_n = (x/4)^n` .

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

`lim_(n-gtoo) |((x/4)^n )^(1/n)| =lim_(n-gtoo) |(x/4)^(n*1/n)|`

`=lim_(n-gtoo) |(x/4)^(n/n)|`

` =lim_(n-gtoo) |(x/4)^1|`

`=lim_(n-gtoo) |(x/4)|`

`=|x/4|`

Applying `Llt1` as the condition for absolutely convergent series, we plug-in `L = |x/4|` on `Llt1` . The interval of convergence will be:

`|x/4|lt1`

`-1 ltx/4lt1`

Multiply each part by `4` :

`(-1)*4 ltx/4*4lt1*4`

`-4ltxlt4`

The series may converges when `L =1 ` or `|x/4|=1` . To check on this, we test for convergence at the endpoints: `x=-4` and `x=4` 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=-4` on `sum_(n=0)^oo (x/4)^n ` , we get a series:

`sum_(n=0)^oo 1*(-4/4)^n =sum_(n=0)^oo 1*(-1)^n`

It shows that `r=-1` and `|r|= |-1|=1` which satisfies `|r|gt=1` . Thus, the series diverges at the left endpoint.

When we let `x=4` on `sum_(n=0)^oo (x/4)^n` , we get a series:

`sum_(n=0)^oo 1*(4/4)^n =sum_(n=0)^oo 1*(1)^n`

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

Conclusion:

The interval of convergence of `sum_(n=0)^oo (x/4)^n ` is `-4ltxlt4`.

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