Question

Test for convergence or divergence, using each test at least once. Identify which test was used. (a) \(n\) th-Term Test (b) Geometric Series Test (c) \(p\) -Series Test (d) Telescoping Series Test (e) Integral Test (f) Direct Comparison Test (g) Limit Comparison Test $$ \sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) $$

Short answer

The series is convergent.

Step-by-step solution

nth-Term Test

This test states that if the limit as \(n\) approaches infinity of the term \(a_n\) doesn't equal to zero, then the series is divergent. Calculate \( \lim_{{n \to \infty}} \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \). Since this limit equals to zero, the test is inconclusive.

Geometric Series Test

This test applies to series of the form \( a \cdot r^{n-1} \) where \( |r| < 1 \). The series is convergent if \( |r| < 1 \) and divergent otherwise. Our series does not fit this form so we cannot use this test.

p-Series Test

The \(p\)-series test states that the series \( \sum_{n=1}^{\infty} 1/n^p \) is convergent if \( p > 1 \) and divergent if \( p \leq 1 \). Our series does not fit this form so we cannot use this test.

Telescoping Series Test

In telescoping series, most terms cancel out when expanded. After expanding the series, notice that most terms indeed do cancel out, thus it is a telescoping series. The series is convergent and its sum is \(1\).

Integral Test

This test involves taking the integral of the series function and determining if the integral is finite or infinite. The integral of \( 1/(n+1) - 1/(n+2) \) from 1 to infinity is finite, therefore the series is convergent.

Direct Comparison Test

This test involves comparing the series to another series which we know to be convergent or divergent. However, as we have already concluded that the series is convergent, we don't need to use this test.

Limit Comparison Test

This test involves finding the limit of \( a_n/b_n \), where \( a_n \) is the given series and \( b_n \) is a series known to be convergent or divergent. However, as we have already concluded that the series is convergent, we don't need to use this test.

Key concepts

nth-Term Test

The nth-Term Test is a fundamental concept used in determining the convergence of a series. It states that if the limit of the sequence's terms does not approach zero as n approaches infinity, the series must diverge. In mathematical terms, if \( \lim_{{n \to \infty}} a_n eq 0 \) then the series \( \sum_{n=1}^\infty a_n \) diverges. It's important to note that if the limit is zero, this test is inconclusive; the series may still converge or diverge.

When we look at an expression like \( \frac{1}{n+1} - \frac{1}{n+2} \) and take the limit as n approaches infinity, we find that it does indeed equal zero. However, this doesn't immediately tell us anything about the overall convergence of the series, so we must use additional tests to determine its behavior.

Geometric Series Test

The Geometric Series Test is a specific test for a series that has a constant ratio between consecutive terms, known as a geometric series. Such a series can be expressed in the form \( a \cdot r^{n-1} \) where 'a' is the first term and 'r' is the common ratio. For convergence, this test simply requires that the absolute value of the common ratio is less than one, \( |r| < 1 \).

If a series does not have a geometric form, this test cannot be applied, which is why it wasn't suitable for the provided series in the exercise. It's a quick and effective tool but limited to the very specific format of geometric progressions.

p-Series Test

The p-Series Test is used to determine the convergence of series in the form \( \sum_{n=1}^\infty \frac{1}{n^p} \) where 'p' is a positive constant. If 'p' is greater than 1, the series is convergent; if 'p' is less than or equal to 1, it is divergent. This test hinges on the understanding that a larger exponent in the denominator will cause terms to approach zero more rapidly, increasing the likelihood of convergence.

The series in our exercise does not fit the p-series format due to the differing denominators and therefore this test is not the right match for determining its convergence.

Telescoping Series Test

In a Telescoping Series, we can observe that many terms cancel each other out as the series progresses. This pattern leads to a series where only a few terms remain after cancellation. To apply the Telescoping Series Test, we expand several terms of the series and analyze the pattern of cancellation. If the remaining few terms form a sequence that has a finite limit, the series is convergent.

For the given exercise, telescoping was evident as subsequent terms negate each other, leaving behind a simple expression whose sum can be found to be a finite number, hence demonstrating convergence of the series.

Integral Test

The Integral Test can be used to establish the convergence of series by comparing it to an improper integral. If \( f(n) = a_n \) is continuous, positive, and decreasing for all \( n \) large enough, then the series and the integral \( \int_{N}^{\infty} f(x) dx \) will both converge or both diverge. You integrate the function from some large 'N' to infinity and if the integral is finite, the series is convergent.

In the exercise, after turning the series into an improper integral, it was determined that the integral is finite. Thus, according to the Integral Test, the series converges.

Direct Comparison Test

The Direct Comparison Test is another convergence test that involves comparing our series to another known series. Specifically, if we have two series where \( 0 \leq a_n \leq b_n \) for all n, and \( \sum b_n \) is a convergent series, then \( \sum a_n \) will also converge. Conversely, if \( \sum b_n \) is divergent, and \( a_n \geq b_n \) for all n, then \( \sum a_n \) will also diverge.

Because we determined that our original series is convergent, there was no need to apply the Direct Comparison Test in the exercise. If the status of our original series was unknown, we would look for a related series that is 'easier' to analyze and use it for comparison.

Limit Comparison Test

Finally, the Limit Comparison Test is similar to the Direct Comparison Test but is used when the terms of the series being compared are not as easily bounded. Instead, it considers the limit of the ratio of the terms of the two series, \( \lim_{n \to \infty} \frac{a_n}{b_n} \) where \( a_n \) is the nth term of the series in question and \( b_n \) is the nth term of a known comparison series. If this limit is positive and finite, and the comparison series is known to converge, then the original series also converges. If the comparison series diverges, then the original series diverges, too.

As with Direct Comparison, since the series was already shown to be convergent, the Limit Comparison Test was not needed for the exercise.