# 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

The terms of a series are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning.
$\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$
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Mayme

Consider the given terms for the infinite series. Find the ratio of the $\left(n+1{\right)}^{th}$ and ${n}^{th}$ term. Use the ratio test to evaluate the nature of the series.
${a}_{1}=\frac{1}{5}$
Now,
$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\mathrm{cos}n+1}{n}$
Let $L=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n+1}}{{a}_{n}}$
Further, simplify using the ratio test. Hence, from the ratio test, the series converges.
So, $L=\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{cos}n+1}{n}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{\mathrm{cos}n+1}{n}}{\frac{n}{n}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{\mathrm{cos}n}{n}+\frac{1}{n}}{1}$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{cos}n}{n}+\frac{1}{n}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{cos}n}{n}\right)+\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}\right)$
$0+\frac{1}{\mathrm{\infty }}$ [From squeeze theorem $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\mathrm{cos}n}{n}\right)=0$]
$0+0$
So, $L=0$
Thus, $L<1$
Hence, $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ converges.

Jeffrey Jordon