Question

Use the test of your choice to determine whether the following series converge. $$\frac{1}{2^{2}}+\frac{2}{3^{2}}+\frac{3}{4^{2}}+\cdots$$

Short answer

Answer: The series converges.

Step-by-step solution

Identify the sequences

First, we have to identify the terms of the sequence. The given series is: $$\frac{1}{2^{2}}+\frac{2}{3^{2}}+\frac{3}{4^{2}}+\cdots$$ This can be written as: $$\sum_{n=1}^{\infty} \frac{n}{(n+1)^2}$$

Set up a comparison

We need to find a convergent or divergent sequence to compare our sequence with. Let's compare it with the convergent sequence \(\frac{1}{n(n+1)}\). Therefore, the comparison is: $$\frac{n}{(n+1)^2} \text{ and } \frac{1}{n(n+1)}$$

Check inequalities

Now, let's check if the following inequality holds for all \(n \ge 1\): $$0 \le \frac{1}{n(n+1)} \le \frac{n}{(n+1)^2}$$ Since the numerator is less than or equal to the denominator for both fractions, the inequality holds.

Confirm convergence of the comparison sequence

Since we are using the Comparison Test, we need to confirm convergence of the compared sequence. The sum of the sequence \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) can be found by partial fraction decomposition: $$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$$ Multiplying both sides by \(n(n+1)\), we get: $$1 = A(n+1) + Bn$$ Comparing coefficients, we have: - For \(A\): \(1 = A(1+1)\) ⇒ \(A = \frac{1}{2}\) - For \(B\): \(1 = B(1)\) ⇒ \(B = 1\) Substituting values of \(A\) and \(B\) in the partial fraction decomposition, we get: $$\frac{1}{n(n+1)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ Now we can compute the sum of the series: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2} \cdot \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ Now using the telescoping method, the series will collapse and: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2} (1)$$ Since the compared sequence converges, we can conclude that our original sequence converges as well.

Conclusion

By the Comparison Test, the original series $$\sum_{n=1}^{\infty} \frac{n}{(n+1)^2}$$ converges.

Key concepts

Comparison Test

The Comparison Test is a popular method for determining the convergence of a series by comparing it to another series whose behavior is already known. The idea is to find a series that you either know converges or diverges and then compare it to your original series. This is helpful because if your known series converges and is greater than the original series term-by-term, then the original series also converges.
If your comparison series diverges and is less than the original series term-by-term, then the original series also diverges.

To apply the Comparison Test correctly, it's important to:

• Select a suitable series for comparison.
• Ensure the inequalities among terms are satisfied.
• Prove convergence or divergence of the comparison series.

In the exercise, the original series \( \sum_{n=1}^{\infty} \frac{n}{(n+1)^2} \) was compared to \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \), a series known to converge. This method was chosen because the terms of \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \) are simpler and lead easily to a telescoping series.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex rational expressions into simpler fractions, which are easier to work with, especially when evaluating limits or sums in a series. This method shines when dealing with rational expressions of polynomials like \( \frac{1}{n(n+1)} \), which appear frequently in series analysis.

The process involves:

• Expressing the fraction as a sum of simpler fractions with unknown coefficients.
• Multiplying through by the common denominator to clear fractions.
• Solving for the unknown coefficients by comparing coefficients or by substitution.

In our exercise, we had \( \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \), and by solving, found \( A = \frac{1}{2} \) and \( B = 1 \).
This decomposition facilitated the use of telescoping series to prove convergence.

Telescoping Series

A Telescoping Series is a series in which many terms cancel out with preceding or succeeding terms, making the sum much simpler to compute. This can occur when the series is represented in a form that reveals this cancellation.
The main steps involved in solving a telescoping series include:

• Writing the series in a form where cancellation can be easily identified.
• Identifying and performing the cancellation of terms.
• Evaluating the few terms that do not cancel, often as a limit as the number of terms goes to infinity.

In the provided solution, the series \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \) is telescopic.
After several terms cancel out, we're left with a simple sum that is often just the first few terms, leading quickly to a conclusion about convergence.

This technique, combined with the Comparison Test, confirmed the convergence of the series by simplifying the computation.