Problem 59 - Thomas Calculus

Question

In Exercises \(57 - 82 ,\) use any method to determine whether the series converges or diverges. Give reasons for your answer. $$ \sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n + 2 } - \frac { 1 } { n + 3 } \right) $$

Step-by-Step Solution

Answer

The series converges to \( \frac{1}{3} \).

Step 1: Representing the Series

The series given is \( \sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n + 2 } - \frac { 1 } { n + 3 } \right) \). This series appears to be telescoping because of the subtraction of consecutive terms.

Step 2: Analyzing the Partial Sums

To understand telescoping series, let's consider the partial sum \( S_m \) of the first \( m \) terms. This is given by \( S_m = \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{m+2} - \frac{1}{m+3} \right) \).

Step 3: Simplifying the Partial Sums

In the expression for \( S_m \), notice that the series is telescoping. Most terms cancel out, leaving us with \( S_m = \frac{1}{3} - \frac{1}{m+3} \).

Step 4: Determining Convergence

The value of \( S_m \) as \( m \rightarrow \infty \) is \( \frac{1}{3} - 0 = \frac{1}{3} \), because \( \frac{1}{m+3} \rightarrow 0 \). Since this limit is a finite number, the series converges.

Key Concepts

Telescoping Series

Telescoping series are a special kind of series where most of the terms cancel each other out when we sum them up. This means, when you write out the partial sums or pairs of terms, many terms cancel out, reducing the series to just a few terms. In the series given by the exercise, \[\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n + 2 } - \frac { 1 } { n + 3 } \right)\]you'll notice that each positive term in the sequence pairs with a negative term in a way that they cancel out. This is the defining feature of a telescoping series.

**Why use a telescoping series?**

• They make finding the sum of a series simpler because of the cancellation property.
• Easier to analyze and solve, often reducing complex series to just a few terms.

In our case, once you express the series in terms of partial sums and simplify them, the remaining terms provide an easy path to decipher convergence.

Partial Sums

Partial sums help us understand the behavior of a series by examining its few initial terms. It involves adding up the first few terms to see if there's a pattern or simplification, especially in telescoping series. For the provided telescoping series, \[S_m = \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{m+2} - \frac{1}{m+3} \right)\]notice the pairs of terms cancel each other out, simplifying the expression significantly.

**What can we learn from partial sums?**

• They show how terms combine and cancel out, especially in telescoping series.
• Reveals the core set of terms that determine the series' convergence or divergence.

In our sum, when we applied the cancellation property, we ended up with just two terms, helping us easily identify convergence.

Infinite Series

An infinite series sums an infinite collection of terms from a sequence. The convergence of an infinite series determines whether the series approaches a finite sum as the number of terms increases indefinitely. The series given,\[\sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { n + 2 } - \frac { 1 } { n + 3 } \right)\]is a telescoping infinite series. As the number of terms goes to infinity, the behavior of the series often depends on its simplified form.

In our example, the series converges to a finite number: \[\lim_{{m \to \infty}} S_m = \frac{1}{3}\]

**Understanding infinite series:**

• Infinite series are a key concept in calculus, providing insights into functions and sequences.
• Convergence helps to determine the ultimate value or limit an infinite series approaches.

With telescoping series, this process is often straightforward, making it easier to establish the behavior of infinite series.