Question

Find the radius of convergence and interval of convergence of the series \(\sum\limits_{n = 1}^\infty n !{(2x - 1)^n}\)

Short answer

The radius of convergence is \(R = 0\) and the interval of convergence is \(I = \left\{ {\frac{1}{2}} \right\}\).

Step-by-step solution

Find \(R\)using the Ratio Test.

Let \(x \ne \frac{1}{2}\)

\(\begin{aligned}\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {\frac{{(n + 1)!{{(2x - 1)}^{n + 1}}}}{{n!{{(2x - 1)}^n}}}} \right|\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \frac{{(n + 1)!}}{{n!}} \cdot |2x - 1|\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \frac{{(n + 1) \cdot n!}}{{n!}} \cdot |2x - 1|\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= (n + 1) \cdot |2x - 1| \to \infty \end{aligned}\)

Concept of Ratio test.

The series diverges if the limit is\( > 1\)and converges if it is\( < 1\). When this limit is less than1, the series converges.

By using the Ratio Test, we can conclude that the series diverges for \(x \ne \frac{1}{2}\).

Therefore, \(R = 0\) and the interval of convergence is \({I_c} = \left( {\frac{1}{2},\frac{1}{2}} \right)\).