To find the radius of convergence and interval of convergence of the series: Sum (^infinity down n=1) ((-1)^(n-1))(1/n^2)(x^n). Please provide steps.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the ratio test such that:

`lim_(n-gtoo) |((-1)^n*x^(n+1)/((n+1)^2))/((-1)^(n-1)*x^n/(n^2))|` = `lim_(x-gtoo)|-x*(n^2)/((n+1)^2)| `

You may factor out x since x is not dependent on the limit such that:

`|x|*lim_(x-gtoo)|(n^2)/(n^2 + 2n + 1)| = |x|*1`

The radius of convergence of this power series is R = 1.

By the ratio test, if |x| > 1, the series diverges and if |x| < 1, the series converges.

`|x| lt 1 =gt -1 lt x lt 1`

You need to determine the interval of convergence such that:

x = -1 => `sum_(n=1)^oo` `(-1)^(n-1)(-1)^n/n^2 = sum_(n=1)^oo -1/n^2 = 0`  => the series converges

`(-1)^n*(-1)^n = (-1)^(2n) = 1`

`x = 1 =gt sum_(n=1)^oo (-1)^(n-1)/n^2` , the series diverges

Hence, evaluating the radius of convergence and the interval of convergence yields `R = 1`  and `[-1,1).`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should perform the ratio test to find the radius of convergence and the interval of convergence such that:

`lim_(n-gtoo) |(((-1)^(n)(x^(n+1)))/((n+1)^2+3))/(((-1)^(n-1))((x^n)/(n^2+3)))|`  = `|x|lim_(n-gtoo) (n^2 + 3)/(n^2+2n+4) |x|lim_(n-gtoo) (n^2 + 3)/(n^2+2n+4) = |x|*1 = |x|`

Hence, the radius of convergence is R=1.

You need to determine the interval of convergence, hence, you should evaluate the series for x=1 and x=-1 such that:

`x=1 =gt  sum_(n=1)^oo (-1)^(n-1)(1/(n^2+3)) =gt`  the series diverges

`x=-1 =gt sum_(n=1)^oo (-1)^(n-1)(((-1)^n)/(n^2+3))`

`(-1)^n*(-1)^n = (-1)^(2n) = 1`

`sum_(n=1)^oo (1/(n^2+3)) = 0 =gt`  the series converges

Hence, evaluating the radius of convergence yields R=1 and the interval of convergence is [-1,1).

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