Answer
(a) We show that ${a_{n + 1}} = \dfrac{{\left| x \right|}}{{n + 1}}{a_n}$.
(b) We show that the sequence $\left\{ {{a_n}} \right\}$ is eventually strictly decreasing.
(c) We show that the sequence $\left\{ {{a_n}} \right\}$ converges.
Work Step by Step
(a) Let $x \ne 0$ and ${a_n} = \dfrac{{|x{|^n}}}{{n!}}$. Thus,
${a_{n + 1}} = \dfrac{{|x{|^{n + 1}}}}{{\left( {n + 1} \right)!}} = \dfrac{{|x{|^n}\left| x \right|}}{{n!\left( {n + 1} \right)}}$
Since ${a_n} = \dfrac{{|x{|^n}}}{{n!}}$, so
${a_{n + 1}} = \dfrac{{\left| x \right|}}{{n + 1}}{a_n}$
(b) In part (a), we have ${a_{n + 1}} = \dfrac{{\left| x \right|}}{{n + 1}}{a_n}$. So,
$\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{\left| x \right|}}{{n + 1}}$
From the equation above, we see that $\dfrac{{{a_{n + 1}}}}{{{a_n}}} \lt 1$ for $n + 1 \gt \left| x \right|$. Thus,
${a_{n + 1}} \lt {a_n}$, ${\ \ \ }$ for $n \gt \left( {\left| x \right| - 1} \right)$
Hence, the sequence $\left\{ {{a_n}} \right\}$ is eventually strictly decreasing.
(c) We showed in part (b) that $\left\{ {{a_n}} \right\}$ is eventually strictly decreasing.
Since all terms in the sequence are positive, it is bounded below by $M=0$, and hence by 9.2.4 Theorem it converges to a limit $L$ satisfying $L \ge M$. Next, we find $L$.
We have
$L = \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}$
$L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left| x \right|}}{{n + 1}}{a_n} = \left( {\mathop {\lim }\limits_{n \to \infty } \dfrac{{\left| x \right|}}{{n + 1}}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right) = 0\cdot L = 0$
Hence, the sequence $\left\{ {{a_n}} \right\}$ converges.
Since $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{|x{|^n}}}{{n!}} = 0$, by 9.1.6 Theorem $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{x^n}}}{{n!}} = 0$. Thus, Formula (5) is established.
