Question

Two sine series Determine whether the following series converge. a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\)

Short answer

Question: Determine the convergence of the following series: a. \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) b. \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) Answer: - The series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) is not absolutely convergent. - The series \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) is conditionally convergent.

Step-by-step solution

Use the absolute convergence

Check if the series converges absolutely by taking the absolute value of the terms: $$\sum_{k=1}^{\infty} |\sin \frac{1}{k}|$$

Compare with the harmonic series

We know that \(0 \le |\sin x| \le |x|\) for all x in the interval \([0,\pi/2]\). Since \(\frac{1}{k}\) decreases as \(k\) increases, and the series converges in the positive direction, we can compare the given series to the absolute value of the harmonic series: $$0 \le |\sin \frac{1}{k}| \le \frac{1}{k}$$ Notice that the harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k}\) is a well-known divergent series.

Determine the convergence of series a

The comparison test tells us that if the larger series (\(\sum_{k=1}^{\infty} \frac{1}{k}\)) is divergent, then the smaller series in question (\(\sum_{k=1}^{\infty} |\sin \frac{1}{k}|\)) must also be divergent. Since the harmonic series diverges, the series \(\sum_{k=1}^{\infty} |\sin \frac{1}{k}|\) also diverges. Therefore, the original series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) is not absolutely convergent. ##Series b: \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\)##

Apply Dirichlet's Test

We want to see if the series converges conditionally. To apply Dirichlet's Test, we must rewrite the series as a product of two sequences: $$\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k} = \sum_{k=1}^{\infty} a_k b_k$$ where \(a_k = \frac{1}{k}\) and \(b_k = \sin \frac{1}{k}\).

Check for conditions that satisfy Dirichlet's Test

Dirichlet's Test states that if the following two conditions are met, then the series converges: 1. The sequence \(a_k\) is a monotonically decreasing sequence of non-negative terms that converges to 0. 2. The partial sums of the sequence \(b_k\) are bounded. Since \(\frac{1}{k}\) is monotonically decreasing and converges to 0 as \(k\) goes to infinity, condition 1 is satisfied.

Study the partial sums of \(b_k\)

Calculate the partial sums of the sequence \(b_k\): $$S_n = \sum_{k=1}^{n} \sin \frac{1}{k}$$ Since \(-1 \le \sin x \le 1\) for all \(x\), we also have \(-1 \le \sin \frac{1}{k} \le 1\). Therefore, the sequence \(b_k = \sin \frac{1}{k}\) is bounded. Hence, the partial sums of the sequence are also bounded, satisfying condition 2.

Determine the convergence of series b

Because both conditions of Dirichlet's Test are met, we can conclude that the series \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) is conditionally convergent. To summarize: - The series \(\sum_{k=1}^{\infty} \sin \frac{1}{k}\) is not absolutely convergent. - The series \(\sum_{k=1}^{\infty} \frac{1}{k} \sin \frac{1}{k}\) is conditionally convergent.