The terms of a series are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. sum_{n=1}^infty a_n a_1=frac15, a_{n+1}=frac{cos n+1}{n}a_n

Question

The terms of a series are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning.

\sum_{n = 1}^{\infty} a_{n}

a_{1} = \frac{1}{5}, a_{n + 1} = \frac{\cos n + 1}{n} a_{n}

Answer 1

Consider the given terms for the infinite series. Find the ratio of the $(n + 1)^{t h}$ and $n^{t h}$ term. Use the ratio test to evaluate the nature of the series.
$a_{1} = \frac{1}{5}$
Now, $a_{n + 1} = \frac{\cos n + 1}{n} \cdot a_{n}$
$\frac{a_{n + 1}}{a_{n}} = \frac{\cos n + 1}{n}$
Let $L = lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$
Further, simplify using the ratio test. Hence, from the ratio test, the series converges.
So, $L = lim_{n \to \infty} \frac{\cos n + 1}{n}$
$= lim_{n \to \infty} \frac{\frac{\cos n + 1}{n}}{\frac{n}{n}}$
$= lim_{n \to \infty} \frac{\frac{\cos n}{n} + \frac{1}{n}}{1}$
$= lim_{n \to \infty} (\frac{\cos n}{n} + \frac{1}{n})$
$= lim_{n \to \infty} (\frac{\cos n}{n}) + lim_{n \to \infty} (\frac{1}{n})$
$0 + \frac{1}{\infty}$ [From squeeze theorem $lim_{n \to \infty} (\frac{\cos n}{n}) = 0$ ]
$0 + 0$
So, $L = 0$
Thus, $L < 1$
Hence, $\sum_{n = 1}^{\infty} a_{n}$ converges.