Problem 23 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{10 n+1}{n(n+1)(n+2)}\end{equation}

Step-by-Step Solution

Answer

The series converges, verified by telescoping method.

Step 1: Simplify the Series' General Term

The general term is given by \(a_n = \frac{10n + 1}{n(n+1)(n+2)}\). Simplify this term by using partial fraction decomposition. Assume \(\frac{10n + 1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}\). Multiply both sides by the denominator \(n(n+1)(n+2)\) to clear fractions: \[10n + 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)\].Expand the right-hand side, collect like terms, and compare coefficients to solve for \(A\), \(B\), and \(C\).

Step 2: Solve for Coefficients in Partial Fraction Decomposition

After expanding, equating coefficients:- Coefficient of \(n^2\): \(2A + 2B + 1C = 10\)- Coefficient of \(n\): \(3A + 2B + C = 0\)- Constant term: \(2A = 1\)Solve these equations simultaneously to find \(A = \frac{1}{2}\), \(B = -3\), and \(C = \frac{5}{2}\). Thus, the partial fraction decomposition is:\[\frac{10n + 1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{3}{n+1} + \frac{5/2}{n+2}\].Now, rewrite the series using these terms.

Step 3: Rewrite the Series using Partial Fractions

The series can be rewritten as:\[\sum_{n=1}^{\infty} \left( \frac{1/2}{n} - \frac{3}{n+1} + \frac{5/2}{n+2} \right).\]Calculate the first few terms and observe if the series telescopes. This means these sequences can cancel out terms successively as the series progresses.

Step 4: Telescoping Series Analysis

Examine the pattern in the series:- First term: \(\frac{1/2}{1} - \frac{3}{2} + \frac{5/6}\).- Second term: \(\frac{1/2}{2} - \frac{3}{3} + \frac{5/4}\).Continue this process to determine the behavior as \(n\) progresses. As the series expands, most intermediate terms cancel out allowing us to identify the behavior of the series.

Step 5: Determine Convergence or Divergence

Due to cancellation, the remaining sums effectively become dominated by a finite number of terms as \(n\to\infty\), thus the series telescopes. Note that the series converges as the partial sums tend towards a finite limit. Double-check using another method like comparison tests or absolute convergence if necessary, but the telescoping analysis strongly indicates convergence.

Key Concepts

Partial Fraction Decomposition

When faced with a complex rational expression, partial fraction decomposition allows us to break it down into simpler fractions. This is particularly useful when working with series, making them easier to analyze. For example, in the given series \[\sum_{n=1}^{\infty} \frac{10n + 1}{n(n+1)(n+2)}\],we start by expressing it as a sum of simpler fractions:

• Assume \(\frac{10n + 1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}\).
• Clear the fractions by multiplying through by the common denominator \(n(n+1)(n+2)\).

This approach simplifies the series by providing us with manageable components, laying the groundwork for further analysis.

Telescoping Series

A telescoping series is one where successive terms cancel each other out. This happens due to the structure that was revealed via partial fraction decomposition. In the example series, after decomposition:\[\sum_{n=1}^{\infty} \left( \frac{1/2}{n} - \frac{3}{n+1} + \frac{5/2}{n+2} \right),\]many terms cancel out when the sequence is expanded.

• Canceling occurs when, for instance, \(\frac{3}{n+1}\) in one term cancels with the \(-\frac{3}{n+1}\) found in another.
• This mechanism leaves only a few non-canceling terms, simplifying the series evaluation.

By focusing mainly on un-canceled terms, we can deduce the series' behavior more clearly.

Coefficient Comparison

To determine the coefficients (\(A\), \(B\), and \(C\)), we utilize coefficient comparison, which involves equating the expanded polynomial terms.By matching like terms on both sides of the equation:\[10n + 1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1),\]we derive:

• For \(n^2\): \(2A + 2B + C = 10\)
• For \(n\): \(3A + 2B + C = 0\)
• For the constant: \(2A = 1\)

Solving these equations gives the values of \(A\), \(B\), and \(C\), thereby completing the partial fraction decomposition.This comparison is crucial in simplifying complex rational expressions into easier components.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. In this context, our interest is in determining whether such a series converges to a finite value or diverges. The given series:\[\sum_{n=1}^{\infty} \frac{10n + 1}{n(n+1)(n+2)}\]represents an infinite sum due to its limits from 1 to infinity. Assessing its convergence or divergence involves techniques like rearrangement into simpler forms or telescoping.
Infinite series can depict finite limits or grow indefinitely, and understanding their behavior is paramount for their utility in practical applications.

Convergence Tests

To determine the convergence of series, various tests exist that rely on different series characteristics. Some commonly used include:

• Telescoping Series Test: Helpful when a series simplifies significantly through the cancellation of intermediate terms, indicating convergence to a finite sum.
• Comparison Test: Involves comparing with a known series that converges or diverges, relying on inequality.
• Ratio or Root Tests: Used to evaluate series terms' behavior when other tests are less applicable.

For the given exercise, the telescoping nature primarily suggested convergence. By simplifying first, we set the stage for further scrutinizing convergence using any suitable test.