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

Short Answer

Expert verified
The sum is 1.

Step by step solution

01

Understand the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{4}{(4n-3)(4n+1)} \) and need to find its sum. The terms involve the partial fraction decomposition technique.
02

Decompose into Partial Fractions

Consider the general term \( \frac{4}{(4n-3)(4n+1)} \). We write this as a sum of partial fractions: \( \frac{A}{4n-3} + \frac{B}{4n+1} \). Multiply and equate coefficients: \[ 4 = A(4n+1) + B(4n-3) \] This simplifies to: \[ 4 = (4A + 4B)n + (A - 3B) \] From the equation, we equate coefficients, which leads to two equations: 1. \( 4A + 4B = 0 \)2. \( A - 3B = 4 \).
03

Solve the System for Partial Fractions

Let’s solve the two equations. From \( 4A + 4B = 0 \), we get \( A = -B \). Substituting \( A = -B \) into \( A - 3B = 4 \), gives: \[ -B - 3B = 4 \] \[ -4B = 4 \] \[ B = -1 \] Thus \( A = 1 \). The decomposition becomes: \( \frac{1}{4n-3} - \frac{1}{4n+1} \).
04

Write Out the Series with Decomposed Terms

Now the series becomes: \( \sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right) \). Notice how the series is telescoping.
05

Recognize the Telescoping Nature

A telescoping series has terms that cancel out sequentially. Listing a few terms gives:\( \frac{1}{1} - \frac{1}{5} + \frac{1}{5} - \frac{1}{9} + \frac{1}{9} - \frac{1}{13} + \cdots \)Notice that terms like \( \frac{1}{5}, \frac{1}{9}, \cdots \) cancel out, leaving only \( \frac{1}{1} \).
06

Find the Sum

Since all intermediate terms cancel out, only the first term, \( \frac{1}{1} \), remains from the telescoping series. Hence, the sum of the series is 1.

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Key Concepts

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

Partial Fraction Decomposition
Partial Fraction Decomposition is a powerful technique to simplify complex rational functions. It involves expressing a simple rational expression as the sum of two or more simpler fractions. This simplification particularly aids in evaluating complex series.

For instance, consider a complex fraction where the denominator is a product of linear factors. The goal of partial fraction decomposition is to break down a complicated expression into its constituent parts:
  • Identify the factors in the denominator of your rational expression.
  • Express the fraction as a sum of fractions where each factor in the denominator appears separately in the new fractions.
This method helps by reducing the rational expression into manageable parts that are easier to work with.

In the exercise, we decomposed the term \( \frac{4}{(4n-3)(4n+1)} \) into \( \frac{1}{4n-3} - \frac{1}{4n+1} \). This was achieved by solving a system of linear equations which resulted from equating coefficients from the partial fraction decomposition. Understanding this step is crucial for simplifying and eventually finding sums.
Telescoping Series
A Telescoping Series is a kind of series where most terms cancel each other out. After canceling, only a few terms remain, which makes evaluating the sum much simpler. Recognizing this pattern allows for quick identification of what will be left over at the end of the series.

When dealing with a telescoping series:
  • Observe the terms within the series structure to identify a pattern where terms cancel out over successive additions.
  • Typically, what's left after cancelation is the first term and the last term (if the series is finite).
In the solution, after decomposing the original series, we see this property emerge. The series terms from \( \sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right) \) cancel consecutively such that all intermediate terms disappear. This leaves only the first term, which then represents the sum of the entire series.
Mathematical Series
A Mathematical Series sums up a sequence of terms. This series could be finite or infinite. The beauty of an infinite series is that it can converge to a single value, despite having an infinite number of terms.

In mathematical series:
  • The notation \( \sum \) represents a series, summing the terms generated by a certain function over a range.
  • A sequence is determined by nth-term expressions, guiding the creation of individual terms in the series.
The current problem involves an infinite series solved by understanding its decomposed structure through partial fraction decomposition and recognizing the telescoping nature of the series. Knowing how to handle these series involves skills in identifying patterns that simplify the series into solvable components over an infinite range.
Convergence of Series
Convergence of Series is the idea that an infinite series can approach a specific value. For a series to converge, the sum of its terms must approach a particular limit as more terms are added. Otherwise, the series diverges, meaning it fails to settle at any value.

A few hallmarks of convergence include:
  • Checking if the series' partial sums approach a finite limit as the number of terms grows infinitely large.
  • Different types of tests (such as the ratio test, root test, or comparison test) help determine if a series converges.
In the exercise, we intuitively observe the convergence through the telescoping nature. As successive terms cancel each other out, the series ultimately converges to a finite value. This is evident because, as we summed the series: only the very first term influenced the final result, suggesting a convergence to the sum of 1.

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Most popular questions from this chapter

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

Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\).

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

If \(\Sigma a_{n}\) converges and \(\Sigma b_{n}\) diverges, can anything be said about their term-by-term sum \(\Sigma\left(a_{n}+b_{n}\right) ?\) Give reasons for your answer.

Estimate the error if \(P_{4}(x)=1+x+\left(x^{2} / 2\right)+\left(x^{3} / 6\right)+\left(x^{4} / 24\right)\) is used to estimate the value of \(e^{x}\) at \(x=1 / 2\).

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