Question

(a) Determine the domain of\(A(x) = \sum\limits_{n = 1}^\infty {{a_n}} \).

(b) Sketch the \({\bf{k}}\)-th partial sum.

(c) Determine the first 12partial sum along with\(A(x)\)..

Short answer

(a)The domain of \(A(x)\) is \(\mathbb{R}\).

(b)\({S_k} = \sum\limits_{n = 0}^{k - 1} {\frac{{{x^{3n}}}}{{\left( {\prod\limits_{m = 1}^n 3 m} \right)\left( {\prod\limits_{m = 1}^n {(3m - 1)} } \right)}}} \)

(c) \(A(x) = \mathop {\lim }\limits_{k \to \infty } \sum\limits_{n = 0}^{k - 1} {\frac{{{x^{3n}}}}{{\left( {\prod\limits_{m = 1}^n 3 m} \right)\left( {\prod\limits_{m = 1}^n {(3m - 1)} } \right)}}} \)

Step-by-step solution

Concept of Ratio test

For any series\(\sum {{a_n}} \), define\(L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\).

If\(L < 1\), the series is absolutely convergent, if\(L > 1\), the series is divergent and if\(L = 1\)the series may be divergent, conditionally convergent or absolutely convergent.

Apply the Ratio Test

(a)

As given, \(A(x) = \sum\limits_{n = 1}^\infty {{a_n}} \)

We can write the following for \(n \ge 1\)

\(\begin{aligned}{a_{n + 1}} &= \frac{{{x^3}}}{{(3n - 1)3n}} \cdot {a_n}\\\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{{x^3}}}{{(3n - 1)3n}}\end{aligned}\)

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

Since the limit is \(0( < 1)\), irrespective of the value of \(x\). The interval of convergence and the domain of \(A(x)\) is \(\mathbb{R}\)

Hence, the domain of \(A(x)\) is \(\mathbb{R}\)

Draw a graph of the partial sums 

(b)

We can write the \({\bf{k}}\)-th partial sum as follows:

\({S_k} = \sum\limits_{n = 0}^{k - 1} {\frac{{{x^{3n}}}}{{\left( {\prod\limits_{m = 1}^n 3 m} \right)\left( {\prod\limits_{m = 1}^n {(3m - 1)} } \right)}}} \)

Draw a graph of the partial sums using an online graphing calculator as shown in figure 1

Figure 1

Hence, \({S_k} = \sum\limits_{n = 0}^{k - 1} {\frac{{{x^{3n}}}}{{\left( {\prod\limits_{m = 1}^n 3 m} \right)\left( {\prod\limits_{m = 1}^n {(3m - 1)} } \right)}}} \)

Plot the graph of the \({\bf{k}}\)-th partial sum along with \(A(x)\) 

(c)

\(A(x) = \mathop {\lim }\limits_{k \to \infty } \sum\limits_{n = 0}^{k - 1} {\frac{{{x^{3n}}}}{{\left( {\prod\limits_{m = 1}^n 3 m} \right)\left( {\prod\limits_{m = 1}^n {(3m - 1)} } \right)}}} \)

Substitute\(k\)=50 for obtain an accurate graph for \(A(x)\)

Graph for the first 12 partial sums along with \(A(x)\) is shown in figure 2

Figure 2

Hence, \(A(x) = \mathop {\lim }\limits_{k \to \infty } \sum\limits_{n = 0}^{k - 1} {\frac{{{x^{3n}}}}{{\left( {\prod\limits_{m = 1}^n 3 m} \right)\left( {\prod\limits_{m = 1}^n {(3m - 1)} } \right)}}} \)