Problem 62 - Thomas Calculus

Question

Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

Step-by-Step Solution

Answer

Both series converge to 1. The sum of the first \(2n+1\) terms of the first series is slightly more than \(1-\frac{1}{n+1}\).

Step 1: Analyze the First Series

The given series is: \[ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots \]Each pair of terms subtracts out, resulting in: \[ \sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right) \]The first \(2n\) terms of this series will thus sum to \( \frac{1}{1} - \frac{1}{n+1} \).

Step 2: Calculate the Sum for the Second Series

The second series is:\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n \cdot (n+1)} \]This is known as a telescoping series:\[ \sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right) \]Therefore, the sum of the first \(n\) terms is also \( \frac{1}{1} - \frac{1}{n+1} \).

Step 3: Show Equivalence of Both Sums

Both series are telescoping and evaluate to the same expression: \[ \sum = \frac{1}{1} - \frac{1}{n+1} \]Thus, the sum of the first \(2n\) terms of the first series is the same as the sum of the first \(n\) terms of the second series.

Step 4: Series Convergence Analysis

As \(n\) approaches infinity, \( \frac{1}{n+1} \to 0\). Therefore, both series converge to 1 as their infinite sum value.

Step 5: Sum of First \((2n+1)\) Terms of First Series

When including one additional term (e.g., the \((2n+1)\)th term), the sequence in the series introduces an incomplete pair, hence the sum would be slightly greater than \( \frac{1}{1} - \frac{1}{n+1} \).

Key Concepts

Sequence and Series

A sequence is an ordered list of numbers, while a series is the sum of the terms of a sequence. These mathematical concepts help us understand patterns and structures in numbers.

• **Sequence:** A sequence is a list of numbers arranged in a specific order. An example of a sequence is \(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\).
• **Series:** When you add the terms of a sequence together, you form a series. For instance, if you take the sequence mentioned above and add its terms, you get the series \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots\).

In the problem illustration, we're focusing on specific types of sequences and series, where the terms follow a distinct pattern. The first series in the exercise forms a specific type of sequence by alternating between positive and negative terms. In these series, a lot cancels out when summed, achieving a core concept known as telescoping. This results in a simplified expression by cancelling intermediate terms, important for understanding more complex infinite series.

Convergence of Series

The convergence of a series indicates whether summing its infinite terms approaches a finite value. Not all series converge, which makes this an essential topic to understand in mathematics.
To determine convergence:

• Examine the behavior of a series' partial sums. As you add more terms, check if the cumulative sum approaches a fixed number.
• If a series converges, it closes in on a particular value as you sum infinitely many terms; otherwise, it diverges and grows indefinitely or oscillates.

In the exercise, both discussed series have a sum of \(1\) as \(n\) tends toward infinity. This is calculated by observing that \(\frac{1}{n+1}\) approaches \(0\), resulting in the infinite sum reducing neatly to \(1\). Understanding this convergence is crucial for mastering the manipulation and application of telescoping series.

Infinite Series

An infinite series is a series that includes endless terms added sequentially. This concept is foundational in calculus and real analysis where series are used to approximate functions, study symmetry, and solve differential equations.

• **Simplicity through Telescoping:** The telescoping nature of a series reduces complexity by canceling terms, helping in calculating infinite sums more easily. Ultimately, it allows one to understand and find a concise result from what might appear complex initially.

In our example, recognizing the pattern of alternating terms allows us to simplify the infinite series to a format that evaluates neatly. This confirms that despite an infinite series' daunting appearance, it can sometimes converge to a surprisingly simple result, like \(1\). This ease of settlement reinforces one of the main intrigues of mathematics — discovering simplicity at large scales.