Question

Determine the convergence or divergence of the series using any appropriate test from this chapter. Identify the test used. $$ \sum_{n=1}^{\infty} \frac{\cos n}{2^{n}} $$

Short answer

The series \(\sum_{n=1}^{\infty} \frac{\cos n}{2^{n}}\) converges by the Ratio Test.

Step-by-step solution

Set Up the Ratio-Test

The ratio test is defined as \(\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|\) where \({a_n}\) are the terms of our series. If the limit is less than 1, the series converges; if it is greater than 1, it diverges. If the limit equals 1 or is inconclusive, the test fails to determine convergence. Here, \(a_n = \frac{\cos n}{2^n}\). Accordingly, we need to calculate \(\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} \left|\frac{\frac{\cos (n+1)}{2^{n+1}}}{\frac{\cos n}{2^n}}\right|\).

Simplify and Compute the Limit

We simplify \(\lim_{n \to \infty} \left|\frac{\frac{\cos (n+1)}{2^{n+1}}}{\frac{\cos n}{2^n}}\right|\) to \(\lim_{n \to \infty} \left|\frac{\cos (n+1)}{2 \cdot \cos n}\right|\). Applying the limit, as \(n\) approaches \(\infty\), \(\cos (n+1)\) and \(\cos n\) will oscillate between -1 and 1, giving no particular trend, all the while, \(2^n\) continues to increase without bound. This means the absolute fraction's value will decrease towards 0.

Conclude the Result

Since the limit as evaluated in Step 2 is less than 1 (in this case 0), by the Ratio Test, the series \(\sum_{n=1}^{\infty} \frac{\cos n}{2^{n}}\) is convergent.