Problem 7 - Thomas Calculus

Question

Use the Comparison Test to determine if each series converges or diverges. \begin{equation}\sum_{n=1}^{\infty} \sqrt{\frac{n+4}{n^{4}+4}}\end{equation}

Step-by-Step Solution

Answer

The series converges by the Comparison Test.

Step 1: Identify the Given Series

The series given is \( \sum_{n=1}^{\infty} \sqrt{\frac{n+4}{n^{4}+4}} \). We must determine if it converges or diverges using the Comparison Test.

Step 2: Simplify the Series

Note that \( \sqrt{\frac{n+4}{n^{4}+4}} \approx \sqrt{\frac{n}{n^{4}}} = \frac{1}{n^{3/2}} \) for large values of \( n \), since higher order terms dominate.

Step 3: Choose a Comparison Series

Based on the simplification, we choose the comparison series to be \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \). This is the \( p \)-series with \( p = \frac{3}{2} \), which is known to converge because \( p > 1 \).

Step 4: Apply the Comparison Test

To apply the Comparison Test, compare the terms of the original series with the terms of the comparison series. We have \( \sqrt{\frac{n+4}{n^{4}+4}} \leq \frac{1}{n^{3/2}} \) for all \( n \geq 1 \), since \( n^4 + 4 > n^4 \) implies \( \frac{n+4}{n^{4}+4} < \frac{n+4}{n^{4}} \leq \frac{n}{n^4} \) for large \( n \).

Step 5: Conclude the Result

Since \( 0 \leq \sqrt{\frac{n+4}{n^{4}+4}} \leq \frac{1}{n^{3/2}} \) and the comparison series \( \sum \frac{1}{n^{3/2}} \) converges, by the Comparison Test, the original series \( \sum_{n=1}^{\infty} \sqrt{\frac{n+4}{n^{4}+4}} \) also converges.

Key Concepts

Series Convergence

Understanding series convergence is crucial in calculus and mathematical analysis. When we talk about a series converging, we're essentially exploring whether the infinite sum of a sequence of numbers approaches a finite value as the number of terms goes to infinity. The series in question is given by \( \sum_{n=1}^{\infty} \sqrt{\frac{n+4}{n^{4}+4}} \). Our task is to determine if it converges or diverges.
To analyze this, mathematicians often use various tests like the Comparison Test, Limit Comparison Test, Ratio Test, and others.In simpler terms:

• A series converges if, as you keep adding more and more terms, the sum leaves you closer and closer to some fixed number.
• If a series does not converge, it diverges, meaning the sum grows indefinitely or does not settle on any particular value.

P-Series

P-series are an essential type of series in mathematics, where each term of the series is of the form \( \frac{1}{n^p} \), with \( n \) being a positive integer and \( p \) being a real number. These series are known as p-series, and their behavior is determined by the value of \( p \).Here's how it works:

• If \( p > 1 \), the p-series converges. This means the endless sum settles towards a distinct value.
• If \( p \leq 1 \), the series diverges. The sum doesn't find a resting point as you add more terms.

In our given exercise, we identified that \( \frac{1}{n^{3/2}} \) is a part of the \( p \)-series because \( p = \frac{3}{2} \), and since \( \frac{3}{2} > 1 \), we know this series converges.
Thus, employing a p-series comparison is an efficient way to determine the behavior of complex series.

Limit Comparison Test

The Limit Comparison Test is a handy tool for figuring out whether a series converges, especially when dealing with more complicated sequences.Here is a glance at how you can use the Limit Comparison Test:

• Choose a simpler series that you already know converges or diverges, typically a p-series.
• Calculate the limit of the ratio of the complex series and the simpler series.
• If the limit is a positive finite number, it implies both series either converge or diverge together.

In our exercise, although we used the direct Comparison Test, the Limit Comparison Test would involve comparing the complicated series \( \sqrt{\frac{n+4}{n^{4}+4}} \) and the simpler \( \frac{1}{n^{3/2}} \).
If the test was performed, the limit would reinforce that both series behave similarly, confirming convergence.
This test is useful when direct computation for comparison isn't straightforward.