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.

$\sum _{k=1}^{\infty }\frac{\sin \sqrt{k}}{k^{3/2}}$

Short answer

The series is absolutely convergent.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{k=1}^{\infty }\frac{\sin \sqrt{k}}{k^{3/2}}$

Step 2. Consider the general series.

The general term of the series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{\sin \sqrt{k}}{k^{3/2}}$is given below,

$a_k=\frac{\sin \sqrt{k}}{k^{3/2}}$

The limit comparison test states that for localid="1649154816940" $\sum _{k=1}^{\infty }{a}_k,\sum _{k=1}^{\infty }{b}_k$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.

The terms of the series $\sum _{k=1}^{\infty }\frac{\sin \sqrt{k}}{k^{3/2}}$is positive.

Step 3. Consider the series ∑k=1∞ bk.

The expression$\frac{\sin \sqrt{k}}{k^{3/2}}$ satisfies$\left|\frac{\sin \sqrt{k}}{k^{3/2}}\right|\le \frac{1}{k^{3/2}}$.

The series $\sum _{k=1}^{\infty }{b}_k$is given below,

$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$

The above series is convergent and converges absolutely.

Hence, the given series is absolutely convergent.