Question

Verify that the infinite series diverges. $$ \sum_{n=1}^{\infty} \frac{n !}{2^{n}} $$

Short answer

The infinite series \(\sum_{n=1}^{\infty} \frac{n !}{2^{n}}\) is divergent.

Step-by-step solution

Understand the ratio test

The ratio test determines the convergence of a series by examining the limit as n approaches infinity of the absolute value of the ratio of the (n + 1)th term to the nth term. If this limit, also known as the ratio, is less than 1, then the series is said to be convergent. If the limit is greater than 1 or infinity, the series is divergent. If the limit equals 1, the ratio test is inconclusive.

Apply the Ratio Test

The ratio test involves taking the limit as n approaches infinity of the ratio of the (n+1)th term to the nth term of the series in absolute values. In this case, the series is:\[\sum_{n=1}^{\infty} \frac{n !}{2^{n}}\]The ratio of the (n+1)th term to the nth term is:\[\frac{(n+1)!}{2^{n+1}} \div \frac{n!}{2^{n}} \]Simplify this to: \[\frac{(n+1)!}{2*n!} = \frac{n+1}{2}\]Take the limit as n approaches infinity:\[\lim_{{n \to \infty}} \frac{n+1}{2} = \infty\]This is greater than 1, so the series is divergent according to the ratio test.

Conclusion

Since the limit in the ratio test turned out to be greater than 1, the series \(\sum_{n=1}^{\infty} \frac{n !}{2^{n}}\) is divergent according to the ratio test.