Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series. $$ \sum_{k=1}^{\infty} \frac{2}{(k+2) k} $$

Step-by-Step Solution

Answer

The series converges and its sum is \( \frac{3}{2} \).

Step 1: Write Out the First Few Terms

First, we need to understand the pattern of the series. We can do this by writing out the first few terms:For \( k = 1 \), the term is \( \frac{2}{(1+2) \cdot 1} = \frac{2}{3} \).For \( k = 2 \), the term is \( \frac{2}{(2+2) \cdot 2} = \frac{2}{8} = \frac{1}{4} \).For \( k = 3 \), the term is \( \frac{2}{(3+2) \cdot 3} = \frac{2}{15} \).So the series starts as: \( \sum_{k=1}^{\infty} \left( \frac{2}{3} + \frac{1}{4} + \frac{2}{15} + \cdots \right) \).

Step 2: Simplify the General Term

Notice that the term \( \frac{2}{(k+2)k} \) can be simplified using partial fraction decomposition.Let's set:\[ \frac{2}{(k+2)k} = \frac{A}{k} + \frac{B}{k+2} \]Multiplying through by the common denominator \((k+2)k\), we get:\[ 2 = A(k+2) + Bk \]\[ 2 = Ak + 2A + Bk \]This simplifies to:\[ (A + B)k + 2A = 2 \]Thus, we have the equations:1. \( A + B = 0 \)2. \( 2A = 2 \) From the second equation, \( A = 1 \) and from the first equation, \( B = -1 \). Thus, the series can be written as:\[ \frac{1}{k} - \frac{1}{k+2} \]

Step 3: Identify the Series as Telescoping

The series can be rewritten in the simplified form as a telescoping series:\[ \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+2} \right) \]This means that a lot of terms will cancel out in a large sum. For instance:\[ \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots \]Notice how terms in the middle like \( \frac{1}{3} \) and \( \frac{1}{4} \) cancel when we add more terms.

Step 4: Determine the Remaining Terms

In a telescoping series like this, most terms cancel out, leaving only the first few terms that do not have a matching negative component. The terms left over are from the beginning of the series and the last part that cannot cancel:The sum simplifies to:\[ \frac{1}{1} + \frac{1}{2} - \left( \frac{1}{n+1} + \frac{1}{n+2} \right) \]As \( n \to \infty \), the terms \( \frac{1}{n+1} \) and \( \frac{1}{n+2} \) tend to zero.

Step 5: Calculate the Sum of the Series

The initial terms that do not get cancelled are \( \frac{1}{1} \) and \( \frac{1}{2} \).Therefore, the sum of the series is:\[ 1 + \frac{1}{2} = \frac{3}{2} \].

Key Concepts

Telescoping Series

When dealing with series, identifying certain patterns can be very helpful in analyzing whether a series converges or not. A telescoping series is an excellent example of this. Telescoping refers to how certain terms in the series cancel each other out as the series progresses. This cancellation allows us to sum the series much easier by reducing the number of terms to focus on.

In our example, after using partial fraction decomposition, we noticed that each term in the series can be rewritten to reveal a pattern of cancellation:

• Many terms will have a positive and a negative counterpart elsewhere in the series.
• For the series \( \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+2} \right) \), observe how middle terms cancel each other out when written sequentially.

Ultimately, this means that the first few and last few terms in the series are the most crucial, as almost all other terms disappear. This property simplifies calculating the series sum.

Partial Fraction Decomposition

Partial fraction decomposition is a valuable tool in calculus, especially in dealing with rational expressions. It involves expressing a complex fraction as a sum of simpler fractions. This method is essential for rewriting our series in a simpler form.

In our given series \( \sum_{k=1}^{\infty} \frac{2}{(k+2)k} \), we can simplify each term using partial fraction decomposition:

• By setting \( \frac{2}{(k+2)k} = \frac{A}{k} + \frac{B}{k+2} \) and solving, we find \( A = 1 \) and \( B = -1 \).
• This means each term simplifies to \( \frac{1}{k} - \frac{1}{k+2} \), highlighting a structure suitable for telescoping.

This decomposition makes the series easier to work with and is a preliminary step often used in conjunction with telescoping series to simplify and find convergence or divergence.

Infinite Series

An infinite series is a summation of an infinite sequence of terms. Understanding infinite series is key to working through problems where the series either converges — meaning it approaches a specific value — or diverges — meaning it does not approach a finite value.

In the series \( \sum_{k=1}^{\infty} \frac{2}{(k+2)k} \), the task is to determine whether it converges. Using techniques such as telescoping or partial fraction decomposition, we can break down this series to make convergence determination feasible.

• When dealing with infinite series, convergence relies on whether adding more terms continues to approach a limit.
• In our example, through evaluation, we find that the leftover terms converge to a constant value of \( \frac{3}{2} \).

Infinite series can be complex, but with these techniques, we can carefully approach and resolve them.

Sum of Series

Finding the sum of a series involves determining the total value that the infinite list of terms approaches. This can be straightforward for finite series but requires more sophisticated techniques for infinite series.

In calculating our series \( \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+2} \right) \), the telescoping nature aids in finding the sum:

• Most terms cancel out, specifically those in the middle of the sequence.
• We are left with the tail ends of the sequence, which do not have counterparts to cancel them out \( \frac{1}{1} + \frac{1}{2} - (\frac{1}{n+1} + \frac{1}{n+2}) \).
• As \( n \to \infty \), the remaining terms \( \frac{1}{n+1} + \frac{1}{n+2} \) approach zero.

The result is the sum of the series, which is \( \frac{3}{2} \). Understanding the sum of series not only checks for convergence but gives us the precise value to which the series converges.