Question

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

Short answer

The radius of convergence is \(R = \infty \) and the interval of convergence is \(I = ( - \infty ,\infty )\).

Step-by-step solution

Concept of Radius of convergence.

For a given power series\(\sum\limits_{n = 0}^\infty {{c_n}} {(x - a)^n}\), the series converges for all\(x\). The number\(R\)is called the radius of convergence of the power series. In this case, the radius of convergence\(R = \infty \)and the interval of convergence\(I = ( - \infty ,\infty )\).

Apply Ratio Test to the given equation 

\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n + 1}}}}{{(2n + 1)!}}\)

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^{n + 1}}\frac{{{x^{2n + 3}}}}{{(2n + 3)!}}}}{{{{( - 1)}^n}\frac{{{x^{2n + 1}}}}{{(2n + 1)!}}}}} \right|\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{x^{2n + 3}}} \right|}}{{(2n + 3)!}} \cdot \frac{{(2n + 1)!}}{{\left| {{x^{2n + 1}}} \right|}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{x^{2n + 1}} \cdot {x^2}} \right|}}{{(2n + 1)! \cdot (2n + 2)(2n + 3)}} \cdot \frac{{(2n + 1)!}}{{\left| {{x^{2n + 1}}} \right|}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{x^2}} \right|}}{{(2n + 3)(2n + 2)}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{x^2}} \right|}}{{(2n + 3)(2n + 2)}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{(2n + 3)(2n + 2)}} \cdot {x^2}\)

\(\begin{aligned} = \frac{1}{\infty } \cdot {x^2}\\ = 0\end{aligned}\)

Since the Limit is 0 and hence < 1, irrespective of the value of \(x\)

We can say that the Interval of convergence is \(R\)and Radius of convergence is \(\infty \)