Sequence Limit Analysis

The sequence $\langle x_n \rangle$ is defined by $x_n = \frac{5^n}{n!}$. To determine its behavior, we evaluate its limit as $n$ approaches infinity.

Ratio Calculation for Convergence

Consider the ratio of consecutive terms:

$ \frac{x_{n+1}}{x_n} = \frac{5^{n+1}}{(n+1)!} \times \frac{n!}{5^n} $

Simplifying this expression gives:

$ \frac{x_{n+1}}{x_n} = \frac{5^{n+1}}{5^n} \times \frac{n!}{(n+1)!} = 5 \times \frac{1}{n+1} = \frac{5}{n+1} $

Now, we calculate the limit of this ratio:

$ L = \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = \lim_{n \to \infty} \frac{5}{n+1} = 0 $

Convergence Determination

A limit $L < 1$ for the ratio of consecutive terms typically implies convergence towards 0. More directly, it's a known mathematical result that for any real number $a$, the limit $\lim_{n \to \infty} \frac{a^n}{n!} = 0$. Applying this with $a=5$:

$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{5^n}{n!} = 0 $

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

Final Conclusion

The sequence $\langle x_n \rangle$ converges to 0.