`sum_(n=1)^oo (-1)^nx^n/n` Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.)

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To determine the interval of convergence for the given series: `sum_(n=1)^oo(-1)^nx^n/n` , we may apply Root Test.

In Root test, we determine the limit as:

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

or 

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

The series is  absolutely convergent if it satisfies the Root test condition: L` lt1.`

For the  given series: `sum_(n=1)^oo(-1)^nx^n/n` , we have:

`a_n= (-1)^nx^n/n`

 Then, set-up the limit as:

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

Note: `|(-1)^n|=1` and `1*|x^n| =|x^n|` .

Apply Law of exponents: `(x/y)^n =x^n/y^n` and `(x^n)^m= x^(n*m).`

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

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

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

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

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

Evaluate the limit.

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

                     `= |x|/|1|`

                     `= |x|`

Thus, we have the limit value: `L = |x|` .

 To determine the interval of convergence , we plug-in `L=|x|` on the condition for convergent series: `Llt1` .

`|x| lt 1`

`-1ltxlt1`

 The series may converge at `L=1` . To verify this, we check for the possible convergence at the endpoints of the interval of x.

Using `x=-1` on the series `sum_(n=1)^oo(-1)^nx^n/n` , we get: 

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

                              `=sum_(n=1)^oo 1^n/n`

                              `=sum_(n=1)^oo 1/n`

The `sum_(n=1)^oo 1/n` is in a form of a p-series where `p =1`  satisfies `0ltplt=1` . According to p-series test of convergence, the series `sum_(n=1)^oo 1/n^p` is convergent if `pgt1` and divergent if  `0ltplt=1` .

Thus, the series `sum_(n=1)^oo 1/n `  is divergent at` x=-1` .

Using `x=1` on the series `sum_(n=1)^oo(-1)^nx^n/n` , we get: 

`sum_(n=1)^oo(-1)^n1^n/n=sum_(n=1)^oo(-1*1)^n/n`

                           `=sum_(n=1)^oo (-1)^n/n`

Applying alternating series test on the series `sum (-1)^na_n` where `a_ngt=0` for all `n` , the series is convergent if we have:

1. `lim_(n-gtoo) a_n = 0`

2. `a_n` is a decreasing sequence

In the series `sum_(n=1)^oo (-1)^n/n` or `sum_(n=1)^oo (-1)^n 1/n` , the `a_n= 1/n` is decreasing sequence and `1/ngt=0` for all `n` .

Evaluating the limit:

`lim_(n-gtoo) 1/n = 1/oo = 0`

Thus,  `sum_(n=1)^oo (-1)^n/n` is convergent  at `x=1` .

The series `sum_(n=1)^oo (-1)^n/n` has a positive and negative elements. We check for absolute or conditional convergence.

The `sum a_n` as `sum_(n=1)^oo (-1)^n/n`  is convergent based from alternating series criteria.

 The `sum |a_n|` as `sum_(n=1)^oo |(-1)^n/n|`  or  `sum_(n=1)^oo 1/n` is divergent based form `p-series` criteria.

 Thus, the series `sum_(n=1)^oo (-1)^n/n` is conditionally convergent at `x=1` .

Therefore, the power series `sum_(n=1)^oo(-1)^nx^n/n`  has an interval of convergence: `-1ltxlt=1` .

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