# `sum_(n=0)^oo (-1)^nx^n/(n+1)` Find the radius of convergence of the power series.

`sum_(n=0)^oo (-1)^n(x^n)/(n+1)`

To find the radius of convergence of a series `sum` `a_n` , apply the Ratio Test.

`L = lim_(n->oo) |a_(n+1)/a_n|`

`L=lim_(n->oo) |((-1)^(n+1) x^(n+1)/((n+1)+1))/((-1)^n(x^n)/(n+1))|`

`L=lim_(n->oo) | (-1) * (x^(n+1)/(n+2))/(x^n/(n+1))|`

`L=lim_(n->oo) | (x^(n+1)/(n+2))/(x^n/(n+1))|`

`L= lim_(n->oo) |x^(n+1)/(n+2) * (n+1)/x^n|`

`L=lim_(n->oo) |(x(n+1))/(n+2)|`

`L = |x| lim_(n->oo) |(n+1)/(n+2)|`

`L=|x| * 1`

`L=|x|`

Take...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

`sum_(n=0)^oo (-1)^n(x^n)/(n+1)`

To find the radius of convergence of a series `sum` `a_n` , apply the Ratio Test.

`L = lim_(n->oo) |a_(n+1)/a_n|`

`L=lim_(n->oo) |((-1)^(n+1) x^(n+1)/((n+1)+1))/((-1)^n(x^n)/(n+1))|`

`L=lim_(n->oo) | (-1) * (x^(n+1)/(n+2))/(x^n/(n+1))|`

`L=lim_(n->oo) | (x^(n+1)/(n+2))/(x^n/(n+1))|`

`L= lim_(n->oo) |x^(n+1)/(n+2) * (n+1)/x^n|`

`L=lim_(n->oo) |(x(n+1))/(n+2)|`

`L = |x| lim_(n->oo) |(n+1)/(n+2)|`

`L=|x| * 1`

`L=|x|`

Take note that in Ratio Test, the series converges when L <1.

`Llt1`

`|x|lt1`

By Ratio Test, the series converges when |x|<1.

Therefore, the radius of convergence of the given series is `R=1` .

Approved by eNotes Editorial Team