Question

Determine the radius of convergence of given series.

\(\sum\limits_{n = 0}^\infty {\frac{{{{(n!)}^k}}}{{(kn!)}}} {x^n}\)

Short answer

The radius of convergence of given series is\(R = {k^k}\).

Step-by-step solution

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.

As given, the series is\(\sum\limits_{n = 0}^\infty {\frac{{{{(n!)}^k}}}{{(kn!)}}} {x^n}\)

Find the ratio for the ratio Test

\(\begin{aligned}{a_n} &= \frac{{{{(n!)}^k}}}{{(kn)!}}{x^n}\\{a_{n + 1}} &= \frac{{{{((n + 1)!)}^k}}}{{(k(n + 1))!}}{x^{n + 1}}\end{aligned}\)

\(\begin{aligned}\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{\frac{{{{((n + 1)!)}^k}{x^{n + 1}}}}{{(k(n + 1))!}}}}{{\frac{{{{(n!)}^k}}}{{(kn)!}}{x^n}}}\\\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{{{((n + 1)!)}^k}{x^{n + 1}}}}{{(k(n + 1))!}} \cdot \frac{{(kn)!}}{{{{(n!)}^k}{x^n}}}\\\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{{{((n + 1)n!)}^k}{x^{n + 1}}}}{{(kn + k)!}} \cdot \frac{{(kn)!}}{{{{(n!)}^k}{x^n}}}\\\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{{{(n + 1)}^k}x(kn)!}}{{(kn + k)!}}\\\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{{{(n + 1)}^k}x(kn)!}}{{(kn + k)(kn + k - 1)(kn + k - 2) \cdots (kn + 1)(kn)!}}\\\frac{{{a_{n + 1}}}}{{{a_n}}} &= \frac{{{{(n + 1)}^k}x}}{{(kn + k)(kn + k - 1)(kn + k - 2) \cdots (kn + 1)}}\end{aligned}\)

Do the Ratio Test

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{(n + 1)}^k}x}}{{(kn + k)(kn + k - 1)(kn + k - 2) \cdots (kn + 1)}} \cdot \frac{{\frac{1}{{{n^k}}}}}{{\frac{1}{{{n^k}}}}}} \right|\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {1 + \frac{1}{n}} \right)}^k}|x|}}{{\left( {k + \frac{k}{n}} \right)\left( {k + \frac{k}{n} - \frac{1}{n}} \right)\left( {k + \frac{k}{n} - \frac{2}{n}} \right) \cdots \left( {k + \frac{1}{n}} \right)}}\)

\( = \frac{{{{(1 + 0)}^k}|x|}}{{(k + 0)(k + 0 - 0)(k + 0 - 0) \cdots (k + 0)}}\)

\( = \frac{{|x|}}{{{k^k}}}\)

The series converges if the result of the Ratio Test is \( < 1\) so solve that for \(x\)

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

The interval of convergence is half the width of the interval

\(\begin{aligned}R &= \frac{{{k^k} - \left( { - {k^k}} \right)}}{2}\\R &= \frac{{2{k^k}}}{2}\\R &= {k^k}\end{aligned}\)

Hence, the radius of convergence of given series is\(R = {k^k}\).