Question

Find the radius of convergence and the interval of convergence of the series\(\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n{3^n}}}} \).

Short answer

The radius of convergence is \(R = 3\) and the interval of convergence is\({\rm{I = ( - 3,3)}}\).

Step-by-step solution

The ratio test, alternating series test, p-test and radius of convergence.

Ratio 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.

Alternating series test:

If the alternating series \(\sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} {b^n} = {b_1} - {b_2} + {b_3} - {b_4} + {b_5} - {b_6} + \ldots \) satisfies the condition, \({b_{n + 1}} \le {b_n}\)and\(\mathop {\lim }\limits_{x \to \infty } {b_n} = 0\), then the series is convergent.

p-test:

The p-series is convergent \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) if \(p > 1\)and divergent if\(p < 1\).

Radius of convergence:

“For a given power series\(\sum\limits_{n = 0}^\infty {{c_n}} {(x - a)^n}\), there is a positive number \(R\) such that the series converges if \({\rm{|x - a| < R}}\)and diverges if\(\,{\rm{|x - a| > R}}\). The number\(R\) is called the radius of convergence of the power series.”

use the ratio test, alternating series test, p-test and the definition of radius of convergence for calculation.

The series is\(\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n{3^n}}}} \).

Let\({a_n} = \frac{{{x^n}}}{{n{3^n}}}\)

Then\({a_{n + 1}} = \frac{{{x^{n + 1}}}}{{{3^{n + 1}}(n + 1)}}.\)

Obtain\(\left| {\frac{{{a_{n + 1}}}}{{{a_a}}}} \right|\) to apply the ratio test.

\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{\frac{{{x^{n + 1}}}}{{3n + 1(n + 1)}}}}{{\frac{{{x^n}}}{{{3^n}n}}}}} \right|\)

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

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{x^{n + 1}}}}{{{3^{n + 1}}(n + 1)}}}}{{\frac{{{x^n}}}{{{3^n}}}}}} \right|\\ &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \cdot \frac{{{3^n}n}}{{{x^n}}}} \right|\\ &= \mathop {\lim }\limits_{n \to \infty } \frac{{|x{|^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \cdot \frac{{{3^n}n}}{{|x{|^n}}}\\ &= \mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{3} \cdot \left( {\frac{n}{{n + 1}}} \right)\end{aligned}\)

Divide both the numerator and denominator by\(n\),

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{3} \cdot \left( {\frac{{\frac{n}{n}}}{{\frac{n}{n} + \frac{1}{n}}}} \right)\\ &= \frac{{|x|}}{3}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\frac{1}{n} + 1}}} \right)\end{aligned}\)

Apply the limit and simplify the terms as shown below.

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \frac{{|x|}}{3} \cdot \left( {\frac{1}{{\frac{1}{x} + 1}}} \right)\\ &= \frac{{|x|}}{3} \cdot \left( {\frac{1}{{0 + 1}}} \right)\\ &= \frac{{|x|}}{3}\end{aligned}\)

The series\(\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n{3^n}}}} \) converges whenever\(\frac{{|x|}}{3} < 1\).

Therefore, by the definition of radius of convergence, \(R = 3\).

Obtain the interval of convergence as follows.

\(\begin{aligned}\frac{{|x|}}{3} < 1\\|x| < 3\\ - 3 < x < 3\end{aligned}\)

Suppose\(x = 3\), then the series\(\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n{3^n}}}} \) becomes,

\(\begin{aligned}\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n{3^n}}}} &= \sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{n{3^n}}}} \\ &= \sum\limits_{n = 1}^\infty {\frac{1}{n}} \end{aligned}\)

By the alternating series test,

Let\({b_n} = \frac{1}{n}\)

\(\begin{aligned}n < (n + 1)\\\frac{1}{{(n + 1)}} < \frac{1}{n}\\{b_{n + 1}} < {b_n}\end{aligned}\)

That is, \(bn\) is decreasing.

Obtain the limit\(bn\).

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {b_n} &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\\ &= 0\end{aligned}\)

Thus, the limit \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\).

Therefore, the series\(\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n{3^n}}}} \) converges when\(x = - 3\).

Hence, the interval of convergence is\({\rm{I = ( - 3,3)}}\).