Q. 38 - Calculus

Question

Use any convergence test from this section or the previous section to determine whether the series in Exercises 31–48 converge or diverge. Explain how the series meets the hypotheses of the test you select. 

$\sum _{k=1}^{\infty }\sin \left(\frac{1}{k^2}\right)$

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k^2}\right)$ is Convergent. 

Step 1. Given information

We are given, 

$\sum _{k=1}^{\infty }\sin \left(\frac{1}{k^2}\right)$

Step 2. Checking the Convergence and Divergence

The terms of the series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k^2}\right)$ are positive.

The expression $\sin \left(\frac{1}{k^2}\right)$ satisfies the following inequality,

$\sin \left(\frac{1}{k^2}\right)\le \frac{1}{k^2}$

The series $\sum _{k=1}^{\infty }{b}_k$ for the series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k^2}\right)$ is given by,

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

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

Therefore, the series $\sum _{k=1}^{\infty }{a}_k$ is also convergent.

Hence, the series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k^2}\right)$ is convergent .