Problem 17 - Single Variable Calculus

Question

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{2}\right)}{\sqrt{n^{3}+1}} $$

Step-by-Step Solution

Answer

The given series converges. This is established by taking the absolute value of the given series and comparing it to a convergent \(p\)-series with \(p=\frac{3}{2}\), by using the absolute convergence test and the comparison test.

Step 1: Find the absolute value of the series

To do this, we will take the absolute value of the given function: $$ \left|\frac{\sin\left(\frac{n\pi}{2}\right)}{\sqrt{n^3+1}}\right|\,. $$

Step 2: Simplify the sine function and find a known series to compare

Notice that the sine function $$\sin\left(\frac{n\pi}{2}\right)$$ has a pattern for integer values of n: - When n is even, $$\sin\left(\frac{n\pi}{2}\right) = 0$$ (all zero terms do not contribute - we can ignore them). - When n=1 or n is odd and greater than 1, $$\sin\left(\frac{n\pi}{2}\right) = 1$$ or $$-1$$. Now we need to compare this series with another known series. We can use the \(p\)-series as a comparison point: $$ \sum_{n=1}^{\infty} \frac{1}{n^p} $$ where p > 1 is a convergent series, and if p ≤ 1, it is a divergent series. When n is odd, the absolute value of the given series is: $$ \left|\frac{\sin\left(\frac{n\pi}{2}\right)}{\sqrt{n^3+1}}\right| = \frac{1}{\sqrt{n^3+1}} \leq \frac{1}{n^{\frac{3}{2}}} $$ We have found a series to compare with, where \(p=\frac{3}{2}\). Thus, the comparison series is: $$ \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}} $$

Step 3: Apply the Comparison Test

Since the comparison series we have found is a convergent series (with p = 3/2 > 1), the original series, according to the absolute convergence test and the comparison test, converges as well. To conclude, when n is odd the series converges and when n is even the terms are 0, so for the entire series: $$ \sum_{n=1}^{\infty} \frac{\sin\left(\frac{n\pi}{2}\right)}{\sqrt{n^3+1}} $$ This series converges.

Key Concepts

Comparison Test

The Comparison Test is a powerful tool for determining whether a series converges or diverges. To use this test, you compare the series you are interested in with another series that is already known to converge or diverge. Here’s a simple way to apply the Comparison Test:

• Find a series that looks similar but is simpler to analyze.
• Ensure that each term of your series is less than or equal to each corresponding term of a known convergent series, or greater than or equal to a divergent one.

If your original series's terms are dominated by those of a convergent series, then your series converges as well. In our exercise, we used the Comparison Test to show that the original series is less than a known convergent p-series, allowing us to conclude that our series also converges.

P-Series

P-Series are a specific type of infinite series and are written as:\[\sum_{n=1}^{\infty} \frac{1}{n^p}\]The convergence of a p-series depends on the value of \(p\):

• If \(p > 1\), the p-series converges.
• If \(p \leq 1\), the p-series diverges.

In the given problem, we compared our series to a p-series with \(p = \frac{3}{2}\). Because \(p = \frac{3}{2}\) is greater than 1, the p-series converges. This helps us conclude using the Comparison Test that our original series also converges.

Absolute Convergence

Absolute convergence is a property of an infinite series that indicates more than just convergence. When a series converges absolutely, the series composed of the absolute values of its terms converges as well.Here's what to remember:

• If the series \(\sum |a_n|\) converges, then the series \(\sum a_n\) also converges.
• Absolute convergence is a stronger condition than regular convergence.

In our problem, we determined absolute convergence by considering the absolute value of each term in our series. By showing that it is less than the terms of a known convergent series, we ensured our series converges absolutely.

Trigonometric Functions

Trigonometric functions like sine and cosine often appear in series. Understanding their behavior is crucial when analyzing such series.In this problem:

• The function \(\sin\left(\frac{n\pi}{2}\right)\) exhibits a pattern particular to its arguments being multiples of \(\pi\).
• For even \(n\), the sine function results in zero, making that series term vanish.
• For odd \(n\), the sine function alternates between \(1\) and \(-1\), affecting whether the term is positive or negative.

Recognizing these patterns allows for significant simplification and understanding of the original series, enabling proper analysis and comparison.