Question

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. $$\sum_{n=1}^{\infty} \frac{n}{n^{3}-4}$$

Short answer

The series \(\sum_{n=1}^{\infty} \frac{n}{n^{3}-4}\) converges.

Step-by-step solution

Identify the Series

The series given is \(\sum_{n=1}^{\infty} \frac{n}{n^{3}-4}\). This is a series where the general term \(a_n = \frac{n}{n^{3}-4}\).

Apply the Preliminary Test

Use the preliminary test to determine if the terms tend to zero. Evaluate \(\lim_{n\to\infty} a_n\). \( a_n = \frac{n}{n^{3}-4} = \frac{n}{n^3(1-\frac{4}{n^3})} = \frac{1}{n^2(1-\frac{4}{n^3})}\). As \(n \to \infty\), \(\frac{4}{n^3} \to 0\), so \(a_n \to \frac{1}{n^2} \to 0\). Since the limit is zero, continue with further tests.

Comparison with a p-Series

Compare \(a_n = \frac{n}{n^{3}-4}\) with the p-series \(b_n = \frac{1}{n^2}\). Use the Limit Comparison Test and find the limit: \[\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n \cdot n^2}{n^3-4} = \lim_{n \to \infty} \frac{n^3}{n^3 - 4} = \lim_{n \to \infty} \frac{1}{1 - \frac{4}{n^3}} = 1.\]Since the limit is a positive finite number (1), the series \(\sum_{n=1}^{\infty} \frac{n}{n^{3}-4}\) converges or diverges similarly to \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), which is a convergent p-series with \(p=2>1\).

Conclusion

Based on the Limit Comparison Test with the convergent p-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), the given series \(\sum_{n=1}^{\infty} \frac{n}{n^{3}-4}\) converges.

Key concepts

Preliminary Test

Before diving into more complex tests, we start with the **preliminary test**, also known as the Divergence Test. The idea here is to check if the terms of the series go to zero as the series progresses. For a series \(\backslash sum_{n=1}^{\backslash infty} a_n \), if \(\backslash lim_{n \to \backslash infty} a_n eq 0\), the series must diverge. However, if the limit is zero, the test is inconclusive, and we need further testing.

In our example, \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{n}{n^{3}-4}\), we set \(\backslash a_n = \backslash frac{n}{n^{3}-4}\). We then calculate the limit:
\(\backslash lim_{n\to \backslash infty} \backslash frac{n}{n^{3}-4} = \backslash frac{1}{n^{2}(1- \backslash frac{4}{n^{3}})}\)
As \(\backslash n \to \backslash infty\), the term \(\backslash frac{4}{n^{3}} \backslash to 0\), making \(\backslash frac{1}{n^{2}} \to 0\). Since the limit is zero, we cannot conclude anything here, so we proceed to other tests.

Limit Comparison Test

The **Limit Comparison Test** is useful when we want to compare our series with another series that we know converges or diverges. If the comparison yields a positive finite limit, then both series will either converge or diverge together. For our series \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{n}{n^{3}- 4}\), we compare it with the p-series \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{1}{n^{2}}\).

We compute the limit:
\(\backslash lim_{n \to \backslash infty} \backslash frac{a_n}{b_n} = \backslash lim_{n \to \backslash infty} \backslash frac{n \backslash cdot n^{2}}{n^{3} - 4} = \backslash lim_{n \to \backslash infty} \backslash frac{n^{3}}{n^{3} - 4} = \backslash lim_{n \to \backslash infty} \backslash frac{1}{1- \backslash frac{4}{n^3}} = 1\).
Here, the limit is 1, which is a positive finite number. This comparison tells us that our series \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{n}{n^{3}- 4}\) will behave similarly to \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{1}{n^{2}}\). Since we know the latter is convergent, our series must also converge.

p-series

A **p-series** is a series in the form \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{1}{n^{p}}\), where \(p\) is a positive constant. The convergence of a p-series depends on the value of \(p\):

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

A classic example is the series \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{1}{n^{2}}\), which converges because \(p=2\) and \(2>1\).

In our case, we used the p-series \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{1}{n^{2}}\) for comparison because it has a known behavior and straightforward limit properties. This comparison, checked via the Limit Comparison Test, confirmed that our original series \(\backslash sum_{n=1}^{\backslash infty} \backslash frac{n}{n^{3}-4}\) converges as well, due to its similarity to a convergent p-series with \(p = 2\).