Question

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{4 n}}{(2 n+1) !} $$

Short answer

Since the limit is 0 which is less than 1, the given series converges according to the Ratio Test.

Step-by-step solution

Derive the General Term

In order to proceed with the ratio test, we need to find a general term for the sequence generated by this series. The general term of our series is \(a_n = \frac{(-1)^{n} 2^{4 n}}{(2 n+1) !}\)

Set Up the Ratio

The ratio test looks at the absolute value of the ratio of the \(n + 1\) term to the \(n\)th term. So we need to set up the ratio: \[ R = \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} 2^{4 (n+1)}}{(2 (n+1)+1) !} / \frac{(-1)^{n} 2^{4 n}}{(2 n+1) !} \right| \]

Simplify the Ratio

After a bit of simplification, this ratio simplifies to: \[ R = \frac{16}{2n+3} \]

Apply the Ratio Test

The ratio test says that an infinite series converges if the limit of this ratio as \(n\) approaches infinity is less than 1. Computing the limit gives: \[ \lim_{n \to \infty} \frac{16}{2n+3} = 0 \]