Question

Use the preliminary test to decide whether the following series are divergent or require further testing. Careful: Do not say that a series is convergent; the preliminary test cannot decide this. $$\sum_{n=1}^{\infty} \frac{(-1)^{n} n}{\sqrt{n^{3}+1}}$$

Short answer

Further testing is required.

Step-by-step solution

Understand the Preliminary Test

The preliminary test (or the nth-term test for divergence) states that if the limit of the nth term of the series as n approaches infinity is not zero, the series diverges.

Identify the nth Term

In the given series, the nth term is \(a_n = \frac{(-1)^{n} n}{\sqrt{n^{3}+1}}\).

Find the Limit of the nth Term

Evaluate the limit of \(a_n\) as \(n \rightarrow \infty\): \[\lim_{n \rightarrow \infty} \frac{(-1)^{n} n}{\sqrt{n^{3}+1}}\]\. To find this limit, observe that for large values of \(n\), the dominant term in \(\sqrt{n^{3}+1}\) is \(n^{3/2}\). Thus, \[\r\frac{n}{\sqrt{n^{3}+1}} \approx \frac{n}{\sqrt{n^{3}}} = \frac{n}{n^{3/2}} = \frac{n}{n^{1.5}} = \frac{1}{n^{0.5}} = \frac{1}{\sqrt{n}}\]\.

Simplify the Limit

Continue simplifying the limit: \[\r\frac{(-1)^{n}}{\sqrt{n}}\]\ as \(n \rightarrow \infty\). Since \(\frac{1}{\sqrt{n}}\) approaches 0 as \(n \rightarrow \infty\), \[\r\frac{(-1)^{n}}{\sqrt{n}} \rightarrow 0\]\.

Conclusion Based on the Preliminary Test

Since \[\r\lim_{n \rightarrow \infty} \frac{(-1)^{n} n}{\sqrt{n^{3}+1}} = 0\]\, the preliminary test alone cannot determine whether the series converges or diverges. Further testing is required.

Key concepts

Series Convergence

When studying infinite series, one of the key things we want to determine is whether the series converges or diverges. Series convergence means that as you add more and more terms of the series, the sum approaches a finite number. For example, the series sum of \(\r\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \text{...}\)\r converges to 1.
To decide on convergence, various tests can be applied. But remember, the main idea is to ensure the partial sums (sums of the first n terms) of the series approach a specific value as n goes to infinity. This is important to know because sometimes the terms of a series go down to zero, but the series itself diverges. Thus, using convergence tests correctly is crucial.

Nth-Term Test

The nth-term test, also known as the Divergence Test, is often the first go-to method to check for divergence in a series. This test simply checks the limit of the nth term as n approaches infinity.
Here's how it works:

• If the limit of the nth term is not zero, the series definitely diverges
• If the limit of the nth term is zero, then the test is inconclusive, and you need further testing to determine convergence

For our exercise, the nth-term test was used:
The nth term of the series \(\r a_{n} = \frac{(-1)^{n} n}{\r\frac{n^{3}+1}}\)\r was found to approach zero as n goes to infinity, which means the test is inconclusive. Hence, more rigorous tests are required to finally decide on the series' behavior.

Limit of a Sequence

In the context of series, understanding the limit of a sequence is essential. The limit of a sequence \(a_n\) as n approaches infinity is the value that \(a_n\) gets closer to as n gets larger.
\r This concept is crucial because every term of the series in question forms a sequence.
To find the limit of the sequence \(a_{n}\) from our exercise:
Identify \(a_{n} = \frac{(-1)^{n} n}{\r\frac{n^{3}+1}}\)\r.
Simplify the expression to get an approximate form for large n, yielding \(a_{n} = \frac{(-1)^{n}}{\r\frac{n}}\)\r.\r Ultimately, the test showed the limit as zero which means that further tests are necessary to conclude the behavior of the series. Understanding and properly applying these principles are key to grasping the different nuances of series convergence.