Question

Determine whether the series is conditionally convergent, absolutely convergent or divergent.

Short answer

The series is absolutely convergent.

Step-by-step solution

Given data

The series is \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}(n + 1){3^n}}}{{{2^{2n + 1}}}}} \).

Concept of convergence and divergence sequence

If\({a_n}\)is a sequence and\(\mathop {\lim }\limits_{n \to \infty } {a_n}\)exists, then the sequence\({a_n}\)is said to be converges otherwise it is diverges.

Simplify the expression

The \(n\)th term of the given series is, \({a_n} = \frac{{{{( - 1)}^{n - 1}}(n + 1){3^n}}}{{{2^{2n + 1}}}}\).

Substitute\(n = n + 1\), then\({a_{n + 1}} = \frac{{{{( - 1)}^n}((n + 1) + 1){3^{n + 1}}}}{{{2^{2(n + 1) + 1}}}}\).

Obtain the limit of \(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\).

\(\begin{array}{c}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{{( - 1)}^n}{{((n + 1) + 1)}^{n + 1}}}}{{{2^{2(n + 1) + 1}}}}}}{{\frac{{{{( - 1)}^{n - 1}}{{(n + 1)}^n}}}{{{2^{2n + 1}}}}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^n}((n + 1) + 1){3^{n + 1}}}}{{{2^{2(n + 1) + 1}}}} \cdot \frac{{{2^{2n + 1}}}}{{{{( - 1)}^{n - 1}}(n + 1){3^n}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{( - 1)(n + 2)3}}{{{2^2}(n + 1)}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{n\left( {1 + \frac{2}{n}} \right)3}}{{{2^2}n\left( {1 + \frac{1}{n}} \right)}}\end{array}\)

Apply limit to the equation

Apply the limit,

\(\begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \frac{{3\left( {1 + \frac{2}{w}} \right)}}{{4\left( {1 + \frac{1}{w}} \right)}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \frac{{3(1 + 0)}}{{4(1 + 0)}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \frac{3}{4}\end{array}\)

Since \(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| < 1\) and by the Ratio Test, the given series is absolutely convergent.

Hence, the series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}(n + 1){3^n}}}{{{2^{2n + 1}}}}} \) is absolutely convergent.