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Chapter 10: Problem 41

Find the sum of each series in Exercises \(41-48\) $$ \sum_{n=1}^{\infty} \frac{4}{(4 n-3)(4 n+1)} $$

Short Answer

Expert verified
The sum of the series is 1.

Step by step solution

01

Simplify the Expression

First, we'll simplify the general term of the series. Our term is \( \frac{4}{(4n-3)(4n+1)} \). Notice that it can be written in a form suitable for partial fraction decomposition.
02

Partial Fraction Decomposition

To decompose \( \frac{4}{(4n-3)(4n+1)} \), express it as:\[\frac{4}{(4n-3)(4n+1)} = \frac{A}{4n-3} + \frac{B}{4n+1}\]Multiply through by \((4n-3)(4n+1)\) and equate coefficients to solve for \(A\) and \(B\) which yields:\[\frac{1}{4n-3} - \frac{1}{4n+1}\]
03

Write the Series with Partial Fractions

Substitute the partial fraction decomposition back into the series:\[\sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right)\]This series is now telescoping, where most terms will cancel out.
04

Identify the Telescoping Nature

Upon expanding the series, notice that for any term \( \frac{1}{4n+1} \), it will cancel with \( \frac{1}{4(n+1)-3} \). This cancellation continues as the series extends.
05

Evaluate the Remaining Terms

After canceling out the terms, you're left with the first non-cancelled part of the sequence. Specifically:\[ \frac{1}{1} \] This is the initial term from \( n=1 \) that does not cancel with any subsequent terms.
06

Conclusion of the Series Sum

Having identified that all other terms in the series cancel out, the sum of the series is:\[1\] This corresponds to the first term that does not get canceled.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
When dealing with complex rational expressions in series, partial fraction decomposition becomes a helpful technique. The goal is to break down a complicated fraction like \( \frac{4}{(4n-3)(4n+1)} \) into simpler terms. This step is crucial because it allows us to rewrite the expression in a format where it becomes easier to identify patterns or cancel out terms.
Here's how it works:
  • First, assume that \( \frac{4}{(4n-3)(4n+1)} \) can be written as the sum of two fractions, \( \frac{A}{4n-3} + \frac{B}{4n+1} \).
  • Next, multiply everything by \((4n-3)(4n+1)\) to get rid of the denominators, leading to an equation in terms of \(n\).
  • Then, solve for the constants \(A\) and \(B\) by equating coefficients on both sides of the equation.
For our exercise, this decomposition resulted in \( \frac{1}{4n-3} - \frac{1}{4n+1} \), setting the stage for further simplification in the series.
Infinite Series
Infinite series are sequences of numbers that continue indefinitely. The sum of such a series is represented as \( \sum_{n=1}^{\infty} a_n \), where each \(a_n\) is a term in the sequence.
Understanding infinite series requires identifying whether the series converges or diverges. This exercise provided an infinite series:
  • \( \frac{4}{(4n-3)(4n+1)} \) was first expressed using partial fractions.
  • It became \( \sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right) \).
This transformation reveals a telescoping series, where consecutive terms cancel each other out, simplifying the process of finding the sum. By its very structure, telescoping helps us see that achieving a finite sum is possible even when dealing with an infinite number of terms.
Series Convergence
Series convergence is fundamental when analyzing infinite series. A convergent series is one where the sum of its terms approaches a finite number as more terms are added. In this exercise, we see a specific case aided by the technique of telescoping.Here鈥檚 how convergence works in a telescoping series:
  • Once the series is expressed as \( \sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right) \), it shows a repetitive cancelation pattern.
  • Each \( \frac{1}{4n+1} \) cancels with the \( \frac{1}{4(n+1)-3} \), leaves only the initial term from the first expansion.
  • After all cancelations, the series converges to the sum of the remaining terms, which in this instance is just \(1\).
Recognizing convergence allows us to conclude that the infinite process resolves to a specific value鈥攅ven as more terms are added.

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Most popular questions from this chapter

Uniqueness of convergent power series a. Show that if two power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(\sum_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) . (Hint: Let \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !) . )\) b. Show that if \(\sum_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n .\)

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} \end{equation}

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so do the following. $$\begin{array}{ll}{\text { a. }} & {\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. }} & {\sum_{n=1}^{\infty} k a_{n}(k \text { any number })}\end{array}$$

Show by example that \(\sum_{n=1}^{\infty} a_{n} b_{n}\) may diverge even if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge.

\begin{equation} \begin{array}{l}{\text { a. Use Taylor's formula with } n=2 \text { to find the quadratic }} \\ {\text { approximation of } f(x)=(1+x)^{k} \text { at } x=0(k \text { a constant) }} \\ {\text { b. If } k=3, \text { for approximately what values of } x \text { in the interval }} \\ {[0,1] \text { will the error in the quadratic approximation be less }} \\ {\text { than } 1 / 100 ?}\end{array} \end{equation}
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