Problem 4 Test the following series for co... [FREE SOLUTION]

Chapter 1: Problem 4

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=0}^{\infty} \frac{n^{2}}{n^{3}+4}$$

Short Answer

Expert verified

The series diverges by the limit comparison test with the harmonic series.

Step by step solution

- Preliminary Test

First, apply the preliminary test. According to this test, if the limit of the term as n approaches infinity is not zero, the series diverges. Compute the limit: \[ \text{lim}_{n \to \rightarrow \frac{n^2}{n^3+4}} \] Divide the numerator and the denominator by $n^3$: \[ \text{lim}_{n \to \rightarrow} \frac{\frac{n^2}{n^3}}{\frac{n^3}{n^3} + \frac{4}{n^3}} = \text{lim}_{n \to \rightarrow} \frac{\frac{1}{n}}{1 + \frac{4}{n^3}} \] As $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$ and $\frac{4}{n^3} \rightarrow 0$. Therefore, \[ \text{lim}_{n \to \rightarrow} \frac{\frac{1}{n}}{1 + \frac{4}{n^3}} = 0 \] Since the limit as $n$ approaches infinity is zero, we proceed to the next step.

- Comparison Test (Comparison with Convergent Series)

Given the series $\sum_{n=0}^{\infty} \frac{n^{2}}{n^{3}+4}$, compare it to a simpler series to see if it converges. Since $\frac{n^{2}}{n^{3}+4}$ has similar growth behavior as $\frac{1}{n}$, let's compare it with $\sum \frac{1}{n}$. Because $\sum \frac{1}{n}$ is the harmonic series and it diverges, this comparison is not helpful to show convergence. So, we consider another test.

- Limit Comparison Test

Apply the limit comparison test. Let's compare $\frac{n^2}{n^3 + 4}$ with $\frac{1}{n}$. Compute the limit: \[ \text{lim}_{n \rightarrow \infty} \frac{\frac{n^2}{n^3 + 4}}{\frac{1}{n}} = \text{lim}_{n \rightarrow \infty} \frac{n^3}{n^3 + 4} \] Simplify the limit: \[ \text{lim}_{n \rightarrow \infty} \frac{n^3}{n^3 + 4} = \text{lim}_{n \rightarrow \infty} \frac{1}{1 + \frac{4}{n^3}} = 1 \] The limit comparison test shows that since $\sum \frac{1}{n}$ (harmonic series) diverges and the limit is a positive finite number (1), $\sum \frac{n^2}{n^3+4}$ also diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Preliminary Test

One of the easiest ways to start when testing a series for convergence or divergence is by applying the preliminary test. This foundational test helps us quickly identify if a series can be immediately classified as divergent. Specifically, this involves examining the limit of the series term as the index approaches infinity. If the limit does not equal zero, the series diverges right away, saving time and effort.

For instance, in our given series $\frac{n^{2}}{n^{3}+4}$, we compute the limit:

\[ \text{lim}_{n \to \rightarrow \frac{n^2}{n^3+4}} \]

By dividing the numerator and the denominator by $n^3$ and simplifying, we get:

\[ \text{lim}_{n \to \rightarrow} \frac{\frac{1}{n}}{1 + \frac{4}{n^3}} = 0 \]

Since the limit is zero, the preliminary test does not conclude divergence. Thus, we must use another test to verify convergence.

Comparison Test

The comparison test is a useful method for determining series convergence by comparing our series with another one that is already known to converge or diverge. This approach involves matching the growth rates of terms between the two series.

In our series $\frac{n^{2}}{n^{3}+4}$, we initially consider comparing it with $ \frac{1}{n} $, the harmonic series, since $ \frac{n^{2}}{n^{3}+4} $ behaves similarly to $ \frac{1}{n} $. However, recognizing that the harmonic series $\frac{1}{n}$ diverges, this comparison doesn't clearly show convergence or divergence for our series.

Limit Comparison Test

When the basic comparison test isn't conclusive, the limit comparison test can be highly efficient. This method involves taking the limit of the ratio of the terms from the two series we are comparing.

For our series, we use $ \frac{1}{n} $ as our comparison series. We calculate:

\[ \text{lim}_{n \to \rightarrow \frac{n^2}{n^3 + 4}}{\frac{1}{n}} = \frac{n^3}{n^3 + 4} \rightarrow 1 \]

Firstly, we note that the harmonic series $ \frac{1}{n} $ diverges. Given that the limit is 1 (a positive finite number), our series $ \frac{n^2}{n^3+4} $ must also diverge based on the limit comparison test since the comparison results in a direct relationship.

Thus, by effectively leveraging the limit comparison test, we conclude that $ \frac{n^2}{n^3+4} $ diverges.

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