Question

Determine the radius of convergence and interval of convergence of a series, \(\sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n{x^n}\).

Short answer

The radius of convergence is \(R = 1\) and the interval of convergence is\({\rm{I = ( - 1,1)}}\).

Step-by-step solution

The ration test.

If \(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\), then the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)is absolutely convergent.

Use the ratio test for calculation.

Let\({a_n} = {( - 1)^n}n{x^n}\).

Then, \({a_{n + 1}} = {( - 1)^{n + 1}}(n + 1){x^{n + 1}}\).

Obtain\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\)

\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{{{( - 1)}^{n + 1}}(n + 1){x^{n + 1}}}}{{{{( - 1)}^n}n{x^n}}}} \right|\)

Take\(\mathop {\lim }\limits_{n \to \infty } \) on both sides,

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^{n + 1}}(n + 1){x^{n + 1}}}}{{{{( - 1)}^n}n{x^n}}}} \right|\\ &= \mathop {\lim }\limits_{n \to \infty } \left| {( - 1)\frac{{(n + 1)x}}{n}} \right|\\ &= \mathop {\lim }\limits_{n \to \infty } \left( {\left( {\frac{n}{n} + \frac{1}{n}} \right)|x|} \right)\\ &= \mathop {\lim }\limits_{n \to \infty } \left( {\left( {1 + \frac{1}{n}} \right)|x|} \right)\end{aligned}\)

Apply the ratio test,

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left( {\left( {1 + \frac{1}{n}} \right)|x|} \right) &= \left( {\left( {1 + \frac{1}{\infty }} \right)|x|} \right)\\ &= ((1 + 0)|x|)\\ &= |x|\end{aligned}\)

The series \(\sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n{x^n}\) converges as\({\rm{|x| < 1}}\).

Therefore, the radius of convergence is\(R = 1\), and the interval is\({\rm{( - 1,1)}}\).

Let\({\rm{x = - 1}}\) be the endpoint of the interval. Then, the series becomes,

\(\begin{aligned}\sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n{x^n} &= \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n{( - 1)^n}\\ &= \sum\limits_{n = 1}^\infty {((} - 1)( - 1){)^n}n\\ &= \sum\limits_{n = 1}^\infty {{1^n}} n\\ &= \sum\limits_{n = 1}^\infty n \end{aligned}\)

Then the series approaches to\(\infty \), which is a divergence series.

Let \({\rm{x = 1}}\) be the endpoint of the interval. Then, the series becomes,

\(\begin{aligned}\sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n{x^n} &= \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n{(1)^n}\\ &= \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} n\end{aligned}\)

The limit does not exist when\(n \to \infty \). Thus, the series diverges.

Therefore, the endpoints cannot consider for the interval of convergence as the limit does not exist.

Hence, the required interval of the convergence is\({\rm{I = ( - 1,1)}}\).