Question

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.

$\sum _{k=1}^{\infty }\frac{\cos k}{k^2}$

Short answer

The series converges absolutely.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{k=1}^{\infty }\frac{\cos k}{k^2}$

Step 2. Consider the general series.

The general term of the series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\cos k}{k^2}$is given below,

$a_k=\frac{\cos k}{k^2}$

The limit comparison test states that for $\sum _{k=1}^{\infty }{a}_k,\sum _{k=1}^{\infty }{b}_k$be two series with positive terms such that $0\le a_k\le b_k$ for every positive integer k. If the series $\sum _{k=1}^{\infty }{b}_k$converges, then the series$\sum _{k=1}^{\infty }{a}_k$converges.

Step 3. Consider the term of the given series as positive.

The given expression satisfies $\left|\frac{\cos k}{k^2}\right|\le \frac{1}{k^2}$.

The series $\sum _{k=1}^{\infty }{b}_k$for the given series isrole="math" localid="1649155585096" $\sum _{k=1}^{\infty }{b}_k\mathit{=}\mathit{\sum }_{k\mathit{=}\mathit{1}}^{\mathit{\infty }}\frac{\mathit{1}}{{\mathit{k}}^{\mathit{2}}}$.

The series $\sum _{k=1}^{\infty }{b}_k\mathit{=}\mathit{\sum }_{k\mathit{=}\mathit{1}}^{\mathit{\infty }}\frac{\mathit{1}}{k^{\mathit{2}}}$ is convergent by p-series test.

The above series is convergent and converges absolutely.

Hence, the given series is absolutely convergent.