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. $$\sum_{n=1}^{\infty} \frac{n-1}{(n+2)(n+3)}$$

Short answer

The series \[\sum_{n=1}^{\infty} \frac{n-1}{(n+2)(n+3)}\] diverges.

Step-by-step solution

Preliminary Test

Check if the limit of the sequence's terms as n approaches infinity is zero. Compute \[\lim_{{n \to \infty}} \frac{n-1}{(n+2)(n+3)}\].

Calculation

Simplify \[\frac{n-1}{(n+2)(n+3)}\] and find the limit:\[\lim_{{n \to \infty}} \frac{n-1}{n^2 + 5n + 6}.\approx \lim_{{n \to \infty}} \frac{n}{n^2} = \lim_{{n \to \infty}} \frac{1}{n} = 0.\] Since the limit is zero, we can proceed to other convergence tests.

Use Partial Fraction Decomposition

Express \[\frac{n-1}{(n+2)(n+3)}\] as the sum of two simpler fractions. Let\[\frac{n-1}{(n+2)(n+3)} = \frac{A}{n+2} + \frac{B}{n+3}.\] Solve for A and B.

Calculation

Multiplying both sides by \[(n+2)(n+3)\], we get:\[n-1 = A(n+3) + B(n+2).\] Expanding and equating coefficients, this results in:\[n-1 = An + 3A + Bn + 2B.\] Matching coefficients of n and the constant term, we get the system:\[A + B = 1\] and\[3A + 2B = -1.\] Solving gives \[A = -3\] and \[B = 4.\]

Rewrite the Series

Substitute A and B back into the partial fractions:\[\frac{n-1}{(n+2)(n+3)} = \frac{-3}{n+2} + \frac{4}{n+3}.\] The series now becomes:\[\sum_{n=1}^{\infty} \left( \frac{-3}{n+2} + \frac{4}{n+3} \right).\]

Test for Convergence

Examine the convergence of each component. Notice that:\[\frac{-3}{n+2}\] and \[\frac{4}{n+3}\] resemble the harmonic series \[\frac{1}{n}\], which is known to diverge.

Calculation

Both series diverge because the harmonic series diverges:\[\frac{-3}{n+2}\] and\[\frac{4}{n+3}\], therefore the combination of these terms results in a divergent series.

Key concepts

Preliminary Test

The Preliminary Test is the first step in determining if a series converges or diverges. It involves checking if the limit of the terms of the sequence goes to zero as n approaches infinity. For the given series \(\frac{n-1}{(n+2)(n+3)}\), we need to calculate the limit \(\text{lim}_{{n \to \text{infinity}}} \frac{n-1}{(n+2)(n+3)}\). Simplifying this expression helps us understand its behavior. By finding that \(\text{lim}_{{n \to \text{infinity}}} \frac{n-1}{n^2+5n+6} = \text{lim}_{{n \to \text{infinity}}} \frac{1}{n} = 0\), we've confirmed that the terms approach zero.

Since the limit is zero, the Preliminary Test allows us to move forward with more detailed convergence tests. This is because the Preliminary Test helps eliminate series that clearly do not converge by checking if the limit is non-zero.

Partial Fraction Decomposition

The next step involves making the series more manageable through Partial Fraction Decomposition. This advanced algebraic technique involves breaking down the complex fraction into simpler parts. For the series \(\frac{n-1}{(n+2)(n+3)}\), we express it as the sum of two simpler fractions: \(\frac{A}{n+2} + \frac{B}{n+3}\).

To find A and B, we multiply both sides by \((n+2)(n+3)\), resulting in \(n-1 = A(n+3) + B(n+2)\). Expanding and equating the coefficients of \('n'\) and the constant terms gives us system equations: \(A + B = 1\) and \(3A + 2B = -1\). Solving these equations, we get A = -3 and B = 4. We then rewrite the original fraction accordingly, making it easier to handle in subsequent convergence tests.

Harmonic Series

After Partial Fraction Decomposition, the series transforms into sums that closely resemble the Harmonic Series. The Harmonic Series is known for its form \(\text{sum}_{n=1}^{\text{infinity}} \frac{1}{n}\), and it is a fundamental example in convergence tests because it diverges.

In our case, the series becomes \(\text{sum}_{n=1}^{\text{infinity}} (\frac{-3}{n+2} + \frac{4}{n+3})\). Here, \(\frac{-3}{n+2}\) and \(\frac{4}{n+3}\) look like harmonic terms. Given that the harmonic series diverges, we suspect that our series might also diverge.

Testing these components against known divergent series gives us crucial insights into the behavior of the original series. This understanding is important because it links our problem to well-known series convergence properties.

Limit Comparison Test

Finally, we can employ the Limit Comparison Test to analyze further. The Limit Comparison Test involves comparing the given series to a known benchmark series to determine convergence or divergence.

In our scenario, we compare \( \text{sum}_{n=1}^{\text{infinity}} (\frac{-3}{n+2} + \frac{4}{n+3})\) to the harmonic series \( \text{sum}_{n=1}^{\text{infinity}} \frac{1}{n}\). We need to check if the ratio of the terms of our series to the harmonic series limits to a positive, finite number.

Analyzing \( \text{lim}_{{n \to \text{infinity}}} \frac{\frac{-3}{n+2} + \frac{4}{n+3}}{\frac{1}{n}}\), we find that it does not converge to a finite and positive limit. Since the harmonic series diverges, and our series' terms mimic its behavior, we conclude our series also diverges.

The Limit Comparison Test provides a structured way to leverage known results to determine the behavior of more complex series. Knowing when and how to use this test is essential for efficiently solving convergence problems.