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. $$\sum_{n=1}^{\infty} \frac{n^{2}-1}{n^{2}+1}$$

Short Answer

Expert verified
The series diverges by the Preliminary Test because the limit of the terms is 1, not 0.

Step by step solution

01

Preliminary Test

First, apply the preliminary test by evaluating the limit of the term as \( n \) approaches infinity. Compute \( \lim_{{n \to \infty}} \frac{{n^2 - 1}}{{n^2 + 1}} \).
02

Simplify the Limit

Simplify the expression under the limit: \( \lim_{{n \to \infty}} \frac{{n^2 - 1}}{{n^2 + 1}} = \lim_{{n \to \infty}} \frac{n^2(1 - \frac{1}{n^2})}{n^2(1 + \frac{1}{n^2})} = \lim_{{n \to \infty}} \frac{1 - \frac{1}{n^2}}{1 + \frac{1}{n^2}} \).
03

Evaluate the Limit

As \( n \) approaches infinity, both \( \frac{1}{n^2} \) terms approach 0. Therefore, \( \lim_{{n \to \infty}} \frac{1 - \frac{1}{n^2}}{1 + \frac{1}{n^2}} = \frac{1 - 0}{1 + 0} = 1 \).
04

Conclusion of Preliminary Test

Since the limit of the terms \( \frac{{n^2 - 1}}{{n^2 + 1}} \) as \( n \) approaches infinity is 1 (not equal to 0), the series \( \sum_{{n=1}}^{{\infty}} \frac{{n^2 - 1}}{{n^2 + 1}} \) diverges by the Preliminary Test.

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Key Concepts

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

Preliminary Test
Before diving into more intricate tests, we start with the preliminary test. This test checks the behavior of the terms in the series as they approach infinity. For the series to potentially converge, the terms must approach zero. If they do not, the series definitely diverges.

For the series \(\bigg\sum_{{n=1}}^{{\text{∞}}} \frac{{n^2 - 1}}{{n^2 + 1}}\), we analyze the limit of \(\frac{{n^2 - 1}}{{n^2 + 1}}\) as \(n \rightarrow \text{∞}\) . We simplify this to:
\(\frac{n^2(1 - \frac{1}{n^2})}{n^2(1 + \frac{1}{n^2})} \).
This further reduces to \(\frac{1 - \frac{1}{n^2}}{1 + \frac{1}{n^2}}\). As \(n\) becomes very large, \( \frac{1}{n^2} \) approaches zero, simplifying our limit to \( \frac{1}{1} = 1 \) .

In this case, since the limit is 1 (not equal to 0), the preliminary test tells us the series \(\bigg\sum_{{n=1}}^{{\text{∞}}} \frac{{n^2 - 1}}{{n^2 + 1}}\) diverges.
Hence, we can conclude that the series will not converge.
Limit Evaluation
Evaluating the limit is a core part of testing the convergence of a series. For our preliminary test, determining the behavior of the terms as \(n\) grows larger is crucial.

For the expression \(\frac{{n^2-1}}{{n^2+1}}\), we start by dividing the numerator and the denominator by \(n^2\). This allows us to simplify the term and make it easier to evaluate the limit.

The simplified form \( \frac{1-\frac{1}{n^2}}{1+\frac{1}{n^2}} \) shows us that as \(n \rightarrow \text{∞}\), both \( \frac{1}{n^2} \) in the numerator and denominator approach zero. Therefore, the entire fraction approaches \( \frac{1}{1} = 1\).

Since this limit is not zero, it indicates that the terms of the series do not diminish to zero, a prerequisite for the series to converge. Therefore, the series diverges. Understanding limit evaluation helps in recognizing how terms behave, simplifying complex fractions, and ultimately deciding convergence or divergence.
Infinite Series
An infinite series is the sum of the terms of an infinite sequence. Understanding whether such a series converges or diverges is fundamental in calculus.

\( \bigg\sum_{{n=1}}^{{\text{∞}}} \frac{{n^2 - 1}}{{n^2 + 1}} \) represents an infinite series where we sum terms indefinitely. For a series to converge, the sum must approach a finite value as we keep adding more terms.

If the sum goes to infinity or fails to settle on a definite amount, the series diverges. One of the first steps is to check individual term behavior, using tests such as the preliminary test and limit evaluation.

In our example, the series diverges because the terms do not diminish to zero. Understanding these concepts, and how to apply related tests, will help you navigate through more complex infinite series evaluations in your studies.

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Most popular questions from this chapter

Assume that the Maclaurin series converges to the function. Show that \(1-\cos x=x^{2} / 2\) with an error less than 0.003 for \(|x|<\frac{1}{2}\).

There are 9 one-digit numbers ( 1 to 9 ), 90 two-digit numbers ( 10 to 99 ). How many three-digit, four-digit, etc., numbers are there? The first 9 terms of the harmonic series \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9}\) are all greater than \(\frac{1}{10}\); similarly consider the next 90 terms, and so on. Thus prove the divergence of the harmonic series by comparison with the series \(\left[\frac{1}{10}+\frac{1}{10}+\cdots\left(9 \text { terms each }=\frac{1}{10}\right)\right]+\left[90 \text { terms each }=\frac{1}{100}\right]+\cdots\) $$ =\frac{9}{10}+\frac{90}{100}+\cdots=\frac{9}{10}+\frac{9}{10}+\cdots $$ The comparison test is really the basic test from which other tests are derived. It is probably the most useful test of all for the experienced mathematician but it is often hard to think of a satisfactory \(m\) series until you have had a good deal of experience with series. Consequently, you will probably not use it as often as the next three tests.

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1-x^{2}}}$$

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}$$

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{e^{n}}{\sqrt{n !}}$$

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