The series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{{\sqrt[3]{{n + 1}}}}} \)
Let\({a_n} = \frac{{{x^n}}}{{\sqrt[3]{n}}}\); ignore the term \({( - 1)^n}\) as it is an absolute value.
Let\({a_{n + 1}} = \frac{{{x^{n + 1}}}}{{\sqrt[3]{{n + 1}}}}\).
The absolute value of \(\frac{{{a_{n + 1}}}}{{{a_n}}}\) is computed as follows,
\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{{x^{n + 1}}}}{{\sqrt[3]{{n + 1}}}} \cdot \frac{{\sqrt[3]{n}}}{{{x^n}}}} \right|\)
\( = \left| {\frac{{\sqrt[3]{n}x}}{{\sqrt[3]{{n + 1}}}}} \right|\)
\( = \left| {\left( {\frac{{\sqrt[3]{n}}}{{\sqrt[3]{{n + 1}}}}} \right)x} \right|\)
\( = \left| {{{\left( {\frac{n}{{n + 1}}} \right)}^{\frac{1}{3}}}x} \right|\)
On further simplification the value of \(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\) becomes,
\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {{{\left( {\frac{n}{{n + 1}},\frac{{\frac{1}{n}}}{{\frac{1}{n}}}} \right)}^{\frac{1}{3}}}x} \right|\)
\( = \left| {{{\left( {\frac{1}{{1 + \frac{1}{n}}}} \right)}^{\frac{1}{3}}}x} \right|\)
Apply the ratio test and simplify the term as shown below,
\(\mathop {\lim }\limits_{n \to \infty } \left| {{{\left( {\frac{1}{{1 + \frac{1}{n}}}} \right)}^{\frac{1}{3}}}x} \right| = \left| {{{\left( {\frac{1}{{1 + \frac{1}{z}}}} \right)}^{\frac{1}{3}}}x} \right|\)
\( = \left| {{1^{\frac{1}{3}}}x} \right|\)
\( = |x|\)
The series\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{{\sqrt[3]{{n + 1}}}}} \)converges as\({\rm{|x| < 1}}\).
Therefore, the radius of convergence is\({\rm{R = 1}}\).
Let\({\rm{x = - 1}}\) be the endpoint of the interval,
\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{{\sqrt[3]{{n + 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{{( - 1)}^n}}}{{\sqrt[3]{n}}}} \)
\( = \sum\limits_{n = 1}^\infty {\frac{{{1^n}}}{{{{(n)}^{\frac{1}{3}}}}}} \)
This is a p-series with\(p = \frac{1}{3}\). That is, the value of\(p\) is less than\(1\).
Thus, the series diverges.
Let\(x = 1\) be the endpoint of the interval,
\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{{\sqrt[3]{{n + 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{{(1)}^n}}}{{\sqrt[3]{n}}}} \)
\( = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{{{(n)}^{\frac{1}{3}}}}}} \)
Consider the series\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{{{(n)}^{\frac{1}{3}}}}}} \), where\({b_n} = \frac{1}{{{{(n)}^{\frac{1}{3}}}}}\).
Then, \({(n)^{\frac{1}{3}}} < {(n + 1)^{\frac{1}{3}}}\).
Thus, \(\frac{1}{{{{(n + 1)}^{\frac{1}{3}}}}} < \frac{1}{{{{(n)}^{\frac{1}{3}}}}}\)
That is, \({b_n} = \frac{1}{{{{(n)}^{\frac{1}{3}}}}}\) is decreasing. …… (1)
Obtain the limit\({b_n} = \frac{1}{{{{(n)}^{\frac{1}{3}}}}}\).
\(\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{{(n)}^{\frac{1}{3}}}}}\)
\( = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } {{(n)}^{\frac{1}{3}}}}}\)
\( = \frac{1}{{{{(\infty )}^{\frac{1}{3}}}}}\)
\( = 0\) (∵1 = 0)
Therefore, the limit is\(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\). …… (2)
From equation (1) and (2), the series\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{{\sqrt[3]{{n + 1}}}}} \) is convergent at the endpoint\(x = 1\) using the alternating series test.
Therefore, the interval of convergence is\({\rm{I = ( - 1,1)}}\).
