Question

Use the special comparison test to find whether the following series converge or diverge. $$\sum_{n=1}^{\infty} \frac{\sqrt{n^{3}+5 n-1}}{n^{2}-\sin n^{3}}$$

Short answer

The series diverges.

Step-by-step solution

Identify the Dominant Terms

Examine the dominant terms in the numerator and denominator. The dominant term in the numerator is \(n^{3/2}\) because it grows fastest as \(n\) increases. The dominant term in the denominator is \(n^2\).

Form a Comparison Series

Compare the given series to a simpler series that has the same dominant terms. We propose comparing to the series \(\sum_{n=1}^{\infty} \frac{n^{3/2}}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}\).

Analyze the Comparison Series

The series \(\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}\) is a p-series with \(p = \frac{1}{2}\), where \(p\leq1\). Thus, the series \(\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}\) diverges.

Apply the Limit Comparison Test

Use the Limit Comparison Test to verify the behavior of the original series. Calculate the limit: \[ \lim_{n \to \infty} \frac{\frac{\sqrt{n^3 + 5n -1}}{n^2 - \sin n^3}}{\frac{n^{3/2}}{n^2}} = \lim_{n \to \infty} \frac{\sqrt{n^3 + 5n -1}}{n^2 - \sin n^3} \cdot \frac{n^2}{n^{3/2}} \] which simplifies to \[ \lim_{n \to \infty} \frac{\sqrt{n^3 + 5n -1}}{n^{3/2}} \]

Simplify the Limit

Simplify the expression inside the limit: \[ \lim_{n \to \infty} \frac{\sqrt{n^3 (1 + \frac{5}{n^2} - \frac{1}{n^3})}}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1 + \frac{5}{n^2} - \frac{1}{n^3}} = 1 \] because as \(n \to \infty\), the terms \(\frac{5}{n^2}\) and \(\frac{1}{n^3}\) approach 0.

Conclude Convergence or Divergence

Since the limit is a positive finite number (1), the behavior of the original series is determined by the comparison series. Therefore, since \(\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}\) diverges, the original series \(\sum_{n=1}^{\infty} \frac{\sqrt{n^{3}+5n-1}}{n^{2}-\sin n^{3}}\) also diverges.

Key concepts

Dominant Terms Identification

When working with series, it's crucial to identify the terms that grow the fastest as \( n \) increases. These are called dominant terms and dictate the behavior of the series at infinity.
In the numerator \( \sqrt{n^3 + 5n - 1} \), the dominant term is \( n^{3/2} \) since \( n^3 \) grows faster than the other terms inside the square root.
In the denominator, \( n^2 - \sin(n^3) \), the dominant term is \( n^2 \), as \( \sin(n^3) \) is bounded between -1 and 1 and doesn't affect the growth rate.
Now our focus on the dominant terms, \( n^{3/2} \) in the numerator and \( n^2 \) in the denominator, simplifies our series' analysis.

p-Series

A p-series is a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Whether a p-series converges or diverges depends on the value of \( p \):

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

In the problem, we compared the given series to the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \), where \( p = 1/2 \). Since \( 1/2 \leq 1 \), this series diverges. This crucial insight tells us about the behavior of the original series.

Series Divergence

A series diverges when its terms do not approach zero or it fails specific convergence tests.
Understanding this concept helps determine the nature of your series.
In our case, by comparing the original series \( \sum_{n=1}^{\infty} \frac{\sqrt{n^3 + 5n - 1}}{n^2 - \sin n^3} \) to the simplified p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \), we see that since the p-series is known to diverge, the limit comparison test will confirm the divergence of the original series too.

Comparison Series

When assessing an unfamiliar series, comparing it to a simpler, well-known series can be helpful. This technique is known as the comparison series method.
The key idea is to identify a series with similar terms whose convergence or divergence is well established.
In this problem, the comparison series used was \( \sum_{n=1}^{\infty} \frac{n^{3/2}}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} \).
This strategic choice simplifies our calculations and verifies if the original series converges or diverges by seeing if the simpler series converges or diverges.
The limit comparison test leveraged this knowledge elegantly to arrive at our final conclusion.