sum_(n=0)^oo (-1)^(n+1)(n+1)x^n Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.)

sum_(n=0)^oo (-1)^(n+1)(n+1)x^n
To determine the interval of convergence, use Ratio Test. The formula of Ratio Test is:
L = lim_(n->oo) |a_(n+1)/a_n|
Applying this formula, L will be:
L = lim_(n->oo) | ((-1)^(n+2)(n+2)x^(n+1))/((-1)^(n+1)(n+1)x^n)|
L=lim_(n->oo) | ((-1)(n+2)x)/(n+1)|
L=|x|*lim_(n->oo)|(n+2)/(n+1)|
 L= |x|*1
L=|x|
Take note that in Ratio Test, the series is convergent when the value of L is less than 1.
L lt 1       :. Series is convergent
So set the L less than 1 to get the values of x in which the series will be convergent.
|x|lt1
-1ltxlt1
So the series converges on the interval -1Next, determine if the series converges or diverges at the end values of the interval.
If x=-1, the series becomes:
sum_(n=0)^oo (-1)^(n+1)(n+1)(-1)^n
= sum_(n=0)^oo (-1)^(2n+1)(n+1)
Then, apply the Series Divergence Test. It states that if the limit of the series is not zero or does not exist, then the series diverges.

lim_(n->oo) a_n !=0    or   lim_(n->oo) = DNE
:.   sum a_n diverges

Taking the limit of the series as n approaches infinity results to:
lim_(n->oo) (-1)^(2n+1) (n+1) = oo
So the series diverges when x=-1.
Moreover, if x=1, the series becomes:
sum_(n=0)^oo (-1)^(n+1)(n+1)(1)^n
=sum_(n=0)^oo (-1)^(n+1)(n+1)
To determine if it is convergent or divergent, apply the Series Divergent Test. Taking the limit of the series as n approaches infinity yields:
lim_(n->oo) (-1)^(n+1)(n+1) = oo
So, the series diverges at x=1.
 
Therefore, the series
sum_(n=0)^oo (-1)^(n+1)(n+1)x^n
is convergent on the interval -1ltxlt1 .