Question

Let $\langle x_n \rangle$ be a sequence which is given by $x_n = \frac{5^n}{n!}$, then

• A. $\langle x_n \rangle$ is not a Cauchy sequence.
• B. $\langle x_n \rangle$ is oscillate.
• C. $\langle x_n \rangle$ is convergent.
• D. $\langle x_n \rangle$ is divergent.

Hint:

Factorials grow much faster than exponentials ($n! \gg a^n$). Therefore, $\lim_{n \to \infty} \frac{a^n}{n!} = 0$ for any real constant $a$.

Answer 1

The Correct Option is C

Step 1: Concept
A sequence of real numbers $\langle x_n \rangle$ is convergent if $\lim_{n \to \infty} x_n = L$ for some finite real number $L$. In complete metric spaces like $\mathbb{R}$, a sequence is convergent if and only if it is a Cauchy sequence.

Step 2: Key Formulas and Approach
To determine the behavior of $x_n = \frac{a^n}{n!}$ (where $a > 0$), we can use D'Alembert's Ratio Test for sequences: If $\lim_{n \to \infty} \left| \frac{x_{n+1}}{x_n} \right| = L$: - If $L < 1$, then $\lim_{n \to \infty} x_n = 0$ (convergent). - If $L > 1$, then $\lim_{n \to \infty} x_n = \infty$ (divergent).

Step 3: Step-by-step Explanation

• Given $x_n = \frac{5^n}{n!}$, all terms $x_n > 0$.

• Compute the ratio $\frac{x_{n+1}}{x_n}$: \[ \frac{x_{n+1}}{x_n} = \frac{\frac{5^{n+1}}{(n+1)!}}{\frac{5^n}{n!}} = \frac{5^{n+1}}{5^n} \cdot \frac{n!}{(n+1)!} \] \[ \frac{x_{n+1}}{x_n} = 5 \cdot \frac{n!}{(n+1)n!} = \frac{5}{n+1} \]
• Take the limit as $n \to \infty$: \[ L = \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = \lim_{n \to \infty} \frac{5}{n+1} = 0 \]
• Since $L = 0 < 1$, by the Ratio Test for sequences, $\lim_{n \to \infty} x_n = 0$.

• Since the limit exists and is a finite real number ($0$), the sequence $\langle x_n \rangle$ is convergent.

• Furthermore, since every convergent sequence in $\mathbb{R}$ is a Cauchy sequence, statement (A) is false. Statements (B) and (D) are also false.

Step 4: Final Answer
The sequence $x_n = \frac{5^n}{n!}$ converges to 0, so it is convergent. Thus, Option (C) is correct.