Question

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

Short answer

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

Step-by-step solution

The ratio test.

If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\), then the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)is absolutely convergent.

Use the ratio test for calculation.

The series is\(\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}}} \).

Let\({a_n} = \frac{{{x^n}}}{{n!}}\).

Then \({a_{n + 1}} = \frac{{{x^{n + 1}}}}{{(n + 1)!}}\)

Obtain \(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\)

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

Simplify the term as shown below,

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

Since \(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\)the radius of convergence is\(R = \infty \)and the interval of convergence is \(I = ( - \infty ,\infty )\).