Question

In the following exercises, use an appropriate test to determine whether the series converges. $$ a_{n}=1 /\left(\begin{array}{c} n+2 \\ n \end{array}\right) \text { where }\left(\begin{array}{l} n \\ k \end{array}\right)=\frac{n !}{k !(n-k) !} $$

Short answer

The series converges, and the sum is 1.

Step-by-step solution

Express Given Terms in Binomial Format

Given the terms of the series are \( a_n = \frac{1}{\binom{n+2}{n}} \). The binomial coefficient \( \binom{n+2}{n} \) can be expanded as \( \frac{(n+2)!}{n! \cdot 2!} \). This simplifies to \( \frac{(n+2)(n+1)}{2} \). Therefore, the term \( a_n = \frac{1}{\frac{(n+2)(n+1)}{2}} = \frac{2}{(n+2)(n+1)} \).

Rewrite the Series

The series we need to analyze is: \( \sum_{n=1}^{\infty} \frac{2}{(n+2)(n+1)} \).

Apply Partial Fraction Decomposition

To make this series easier to analyze, we'll apply partial fraction decomposition. Express \( \frac{2}{(n+2)(n+1)} \) as \( \frac{A}{n+1} + \frac{B}{n+2} \). Solving for \( A \) and \( B \), we get:\[\frac{2}{(n+2)(n+1)} = \frac{1}{n+1} - \frac{1}{n+2}.\]

Recognize the Telescoping Nature of the Series

The series can be written as \( \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \). Recognize that this is a telescoping series, where many intermediate terms will cancel out.

Determine the Sum of the Series

The terms cancel as follows: The series starts with \( \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots \). Most terms cancel, leaving the first negative terms without subsequent values to cancel them. Therefore, the sum of the series is \( 1/2 \).

Conclude with the Result on Convergence

Because the series telescopes to a finite sum, the series converges. The convergence result is that the sum converges to 1.

Key concepts

Binomial Coefficient

The binomial coefficient is a fundamental concept in algebra and combinatorics that appears as part of the solution to the problem. It is represented as \( \binom{n}{k} \) and defined mathematically as \[ \binom{n}{k} = \frac{n!}{k!(n-k)!}. \] This formula provides the number of ways to choose \(k\) elements from a set of \(n\) elements.
This is particularly useful for understanding series and sequences in mathematics. In our exercise, we deal with \( \binom{n+2}{n} \), which simplifies the series terms allowing for further manipulation with ease to reach a conclusion about convergence.

• The factorial function \( n! \) represents the product of all positive integers up to \(n\).
• The numerator \((n+2)!\) takes the product from \(n+2\) down to 1.
• The denominator \(2!\) simplifies this for our case by providing a factor of 2.

By breaking down these components, we can rewrite terms in simpler forms, crucial to applying further algebraic techniques.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a complex rational expression as a sum of simpler fractions. This is particularly useful when dealing with series as it can make the terms easier to handle.
In this particular exercise, we start with the expression \( \frac{2}{(n+2)(n+1)} \). By using partial fraction decomposition, we express this as \( \frac{1}{n+1} - \frac{1}{n+2} \). Here’s how we do it:

• Assume \( \frac{2}{(n+2)(n+1)} = \frac{A}{n+1} + \frac{B}{n+2} \).
• Clear the denominators by multiplying through by \((n+2)(n+1)\).
• This leads to the equality: \(2 = A(n+2) + B(n+1)\).
• By choosing convenient values for \(n\) or equating coefficients, solve for \(A\) and \(B\).

It’s this decomposition that allows us to recognize the series as a telescoping series, greatly simplifying further calculations.

Telescoping Series

A telescoping series is a type of series where successive terms cancel each other out, simplifying the sum. Understanding this nature helps in simplifying complex series calculations.
In our case, we rewrite the series from the partial fraction result: \( \sum_{n=1}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \). Scanning through these terms, notice the pattern:

• The first term of the current expression cancels with the second term of the previous one.
• This leaves only a few initial and final terms from the entire sum.

The basic property of telescoping series is that the series' complexity vanishes, literally and mathematically, allowing us to see that initial terms lead to a clear, finite sum. Recognizing this pattern is key to determining convergence.

Series Convergence

Series convergence is a fundamental concept in calculus and analysis to determine if a sum approaches a fixed value as more terms are added. It answers whether a series will settle to a particular number or diverge indefinitely.
For the series in question, after separating the terms through partial fraction decomposition and identifying its telescoping nature, the result is a clear, finite sum dictated by only the remaining terms: \( \frac{1}{2} \) to infinity but without a matching decrement term for that \( \frac{1}{2} \).
This results from:

• Initial cancellation of terms due to the telescoping characteristic.
• The sum moves closer to its finite value, exemplified by this example: \(( \frac{1}{2} - \lim_{n \to \infty} \frac{1}{n+2} = \frac{1}{2})\).

It is the telescoping action that converges the series sum to 1. Understanding convergence informs us about the behavior of series as terms grow large, and it's fundamental to infer the outcome of infinite processes.