`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.)

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`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 -1<x<1.

Next, 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` .

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