Question

In Exercises $21-30$use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. The comparison test states that for ∑k=1∞ ak and ∑k=1∞ bk be two series with positive terms then,

1. If $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=L$, where $L$is any positive real number, then either both converge or both diverge.
2. If $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=0$and $\sum _{k=1}^{\infty }{b}_k$converges, then $\sum _{k=1}^{\infty }{a}_k$also converges.
3. If$\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\infty$and$\sum _{k=1}^{\infty }{b}_k$diverges, then$\sum _{k=1}^{\infty }{a}_k$also diverges.

Step 3. The terms of the series ∑k=1∞ sin kk2 are positive.

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

$\frac{{\sin }^{2}k}{k^2}\le \frac{1}{k^2}$.

Find the series $\sum _{k=1}^{\infty }{b}_k$for the given series.

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

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

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

Hence, the given series is also convergent.