Q. 30 - Calculus

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.

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

Step-by-Step Solution

Answer

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

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 ∞   a k and ∑ k = 1 ∞   b k 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   k k 2 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.