Problem 6 - Thomas Calculus

Question

In Exercises \(1-6,\) find a formula for the \(n\) th partial sum of each series and use it to find the series' sum if the series converges. $$ \frac{5}{1 \cdot 2}+\frac{5}{2 \cdot 3}+\frac{5}{3 \cdot 4}+\dots+\frac{5}{n(n+1)}+\cdots $$

Step-by-Step Solution

Answer

The series sums to 5.

Step 1: Identify the General Term

The given series is \( \frac{5}{1 \cdot 2} + \frac{5}{2 \cdot 3} + \frac{5}{3 \cdot 4} + \dots \), so the general term of the series can be expressed as \( a_n = \frac{5}{n(n+1)} \).

Step 2: Simplify the General Term

The term \( \frac{5}{n(n+1)} \) can be rewritten using partial fractions. We can express it as \( \frac{A}{n} + \frac{B}{n+1} \). Solving for \( A \) and \( B \), we get \( \frac{5}{n(n+1)} = \frac{5}{n} - \frac{5}{n+1} \).

Step 3: Write the n-th Partial Sum

The nth partial sum of the series is given by \( S_n = \sum_{k=1}^{n} \left( \frac{5}{k} - \frac{5}{k+1} \right) \).

Step 4: Use Telescoping Series Property

Observe that this series is telescoping. Most terms cancel out: \( S_n = \left( \frac{5}{1} - \frac{5}{2} \right) + \left( \frac{5}{2} - \frac{5}{3} \right) + \ldots + \left( \frac{5}{n} - \frac{5}{n+1} \right) \).

Step 5: Find Simplified Partial Sum

After cancellation, what remains is \( S_n = 5 - \frac{5}{n+1} \).

Step 6: Determine Series Sum (If Convergent)

As \( n \to \infty \), \( \frac{5}{n+1} \to 0 \). Therefore, the sum of the series is \( S = \lim_{n \to \infty} S_n = 5 \).

Key Concepts

Partial Fractions

Partial fractions is a technique used to break down complex rational expressions into simpler ones. This method is crucial for dealing with series and integrals involving rational terms. In our exercise, we deal with the term \( \frac{5}{n(n+1)} \). To simplify, it helps to express this term using partial fractions where an expression is broken into the sum of simpler fractions.

• We express \( \frac{5}{n(n+1)} \) as \( \frac{A}{n} + \frac{B}{n+1} \).
• Through algebra, solve for \( A \) and \( B \) by equating and simplifying both sides. This involves matching coefficients or strategically choosing values of \( n \).
• We find \( A = 5 \) and \( B = -5 \), hence, our expression becomes \( \frac{5}{n} - \frac{5}{n+1} \).

This simplification helps further because it makes the series "telescoping," allowing easier summation for convergence analysis.

Convergence of Series

Understanding the convergence of a series is essential in calculus and real analysis. A series is said to converge if its sequence of partial sums approaches a finite limit. In this context, telescoping series play an intriguing role.
A telescoping series is one where consecutive terms cancel each other out. Here, the series \( \sum_{k=1}^{n} \left( \frac{5}{k} - \frac{5}{k+1} \right) \) fits this pattern:

• Each internal element, except for the first and last, cancels with its neighbor.
• This leads to a cleaner, simplified form of the partial sum: \( S_n = 5 - \frac{5}{n+1} \).
• As \( n \to \infty \), \( \frac{5}{n+1} \) approaches zero, causing the series to converge to \( 5 \).

Such telescoping behavior ensures that the convergence is straightforward and elegant — a principle that greatly simplifies the analysis of infinite series.

Partial Sum Formula

The partial sum is a foundational concept in understanding series. It helps us find the sum of a series up to the \( n \)-th term, offering insight into how the series behaves as it approaches infinity.
In our exercise, we derived the partial sum formula as \( S_n = 5 - \frac{5}{n+1} \), using the telescoping property.

• To arrive at this, note the pattern: \( S_n = \left( \frac{5}{1} - \frac{5}{2} \right) + \left( \frac{5}{2} - \frac{5}{3} \right) + \cdots + \left( \frac{5}{n} - \frac{5}{n+1} \right) \).
• Most terms cancel out, leaving \( 5 - \frac{5}{n+1} \) as the sum.
• As \( n \to \infty \), the term \( \frac{5}{n+1} \) decreases to zero, solidifying that the sum of the series converges to \( 5 \).

The partial sum formula not only allows us to determine convergence but also to effectively compute the sum of infinite series, especially in telescoping cases like this.