Problem 15 - Single Variable Calculus

Question

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \sin \frac{(2 n-1) \pi}{2} $$

Step-by-Step Solution

Answer

The given series can be written as an alternating series, with \(a_n = \frac{1}{\sqrt{n}}\) and \(b_n = \sin\frac{(2 n-1) \pi}{2}\). As \(a_n\) is strictly decreasing and its limit as \(n\) approaches infinity is zero, and \(b_n\) is alternating, the Alternating Series Test indicates that the series converges.

Step 1: Identify the terms

The given series can be written as an alternating series by separating the positive and negative terms as follows: \[ \sum_{n=1}^{\infty} a_n b_n \] Where \(a_n = \frac{1}{\sqrt{n}}\) and \(b_n = \sin\frac{(2 n-1) \pi}{2}\). Step 2: Test if the sequence is decreasing

Step 2: Test if decreasing

The sequence \(a_n = \frac{1}{\sqrt{n}}\) is decreasing since the derivative of the function \(f(x) = \frac{1}{\sqrt{x}}\) is: \[ f'(x) = -\frac{1}{2 \sqrt{x^3}} \] Since the derivative is negative for all positive values of \(x\), the sequence is decreasing. Step 3: Test if the limit of the sequence goes to zero

Step 3: Find the limit

Now, let's find the limit of the sequence \(a_n = \frac{1}{\sqrt{n}}\) as \(n\) approaches infinity. \[ \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 \] Step 4: Verify that \(b_n\) is alternating

Step 4: Verify the alternating sequence

Now, let's verify that the sequence \(b_n = \sin\frac{(2 n-1) \pi}{2}\) is alternating: \[ b_n = \sin\frac{\pi}{2} , \sin\frac{3\pi}{2} , \sin\frac{5\pi}{2}, \ldots \] These terms correspond to the sine values of the first, third, fifth, and so on quadrants in a unit circle, which are, respectively, positive, negative, positive, negative, and so on. Step 5: Apply the Alternating Series Test

Step 5: Apply the Alternating Series Test

Since the sequence \(a_n = \frac{1}{\sqrt{n}}\) is strictly decreasing and its limit is zero, and we have verified that the sequence \(b_n\) is alternating, we can apply the Alternating Series Test. By the Alternating Series Test, the given series converges.

Key Concepts

Series Convergence

Understanding series convergence is crucial when dealing with infinite sums. A series \( \sum_{n=1}^\infty a_n \) converges if the sum of its infinite terms approaches a finite number. The convergence of a series depends on the behavior of its terms as the series progresses.

Using specific converging tests can determine the fate of a series, and one such test, applicable here, is the Alternating Series Test. This test concludes the series \( \sum_{n=1}^\infty (-1)^{n-1} a_n \) converges if all of the following are true:

• The \(a_n\) terms are all positive,
• \(a_n\) is a decreasing sequence, and
• The limit of \(a_n\) as \(n\) approaches infinity is zero.

If any of these conditions fail, the test cannot confirm the series' convergence.

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence approach as the index goes to infinity. In mathematical terms, if \( \lim_{n \to \infty} a_n = L \) for a sequence \( (a_n)_{n=1}^\infty \) and a real number \(L\), we say that \(L\) is the limit of the sequence.

Determining the limit is a pivotal part of assessing series convergence. As we've seen, one requirement of the Alternating Series Test is that this limit must be zero. During the application of this test to \( a_n = \frac{1}{\sqrt{n}} \), as \(n\) grows larger, the terms get smaller and approach zero.

Thus, \( \lim_{n \to \infty} a_n = 0 \) demonstrates a necessary condition for convergence in alternating series. Without showing this step, we cannot effectively determine the series' behavior at infinity.

Decreasing Sequence

A sequence is decreasing if each term is less than or equal to the preceding term. Formally, for the sequence \( a_n \), it is decreasing if \( a_{n+1} \leq a_n \) for all \(n\).

The condition of decreasing terms in an alternating series is important because it implies the 'damping effect' needed for convergence; later terms in the series will subtract a smaller and smaller amount from the sum.

To confirm whether our sequence \( a_n = \frac{1}{\sqrt{n}} \) is decreasing, we look at its derivative. Since \( f'(x) = -\frac{1}{2 \sqrt{x^3}} < 0 \) for all positive \(x\), we infer that the sequence is decreasing. This monotonic behavior is essential for applying the Alternating Series Test and thereby proving the convergence of our given series.