Question

Find the radius of convergence and interval of convergence of the series \(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{{(x - 3)}^n}}}{{2n + 1}}\).

Short answer

The radius of convergence is \(R = 1\) and the interval of convergence is \(I = (2,4)\).

Step-by-step solution

Solve the series by using Ratio test.

\(\begin{aligned}\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {\frac{{{{(x - 3)}^{n + 1}}}}{{2(n + 1) + 1}} \cdot \frac{{2n + 1}}{{{{(x - 3)}^n}}}} \right|\\\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {(x - 3) \cdot \frac{{2n + 1}}{{2n + 3}}} \right|\end{aligned}\)

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {(x - 3) \cdot \frac{{2n + 1}}{{2n + 3}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= |x - 3| < 1\end{aligned}\)

After taking the limit as n goes to \(\infty \), by using the ratio test we can observe that the series converges for \(|x - 3| < 1\), i.e., for \(2 < x < 4\)

Substitute the value of \(x\)

When \(x = 2\), the series becomes

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{{( - 1)}^n}}}{{2n + 1}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{2n + 1}}} \), which diverges.

Because it is a broad harmonic series, this series diverges.

When \(x = 4\), the series becomes

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{{(1)}^n}}}{{2n + 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{2n + 1}}} \), which converges.

The series converges, since it is an alternating\(p\)-series with p=1

Hence the resultant Interval of convergence \( = (2,4)\)

Radius of convergence \( = 1\)