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Chapter 3: Problem 20

Find a closed form expression for $$ \prod_{k=1}^{n} \frac{k^{3}-1}{k^{3}+1} $$

Short Answer

Expert verified
The product simplifies to \(\frac{2}{n^3+1}\).

Step by step solution

01

Factorize the Terms in the Product

We begin by factorizing both the numerator and the denominator for each term in the product. Notice that:\[k^3 - 1 = (k-1)(k^2 + k + 1)\]and\[k^3 + 1 = (k+1)(k^2 - k + 1)\]
02

Write the Partial Terms Using the Factors

Express each fraction in the product in terms of these factors:\[\frac{k^3-1}{k^3+1} = \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}\]Note that each of these fractions will have components that cancel out in the larger product expression.
03

Identify the Telescopic Nature of the Product

Observe how the terms cancel when expanded across multiple values of \(k\). Specifically, the \((k-1)\), \((k+1)\), and associated factors simplify when the product is fully expanded due to adjacent terms canceling out. For example, parts of the form \((k^2+k+1)\) and \((k^2-k+1)\) will cancel with similar terms for different \(k\).
04

Simplify the Product

After expanding a few terms, realize that most middle terms cancel out, leaving:\[\frac{1^2 + 1 + 1}{n^3+1}\]The surviving terms are the initial and the final few terms (since their counterparts don't exist to cancel them out).
05

Write the Closed Form Expression

From the cancellation, the closed form expression for the product can be simplified as:\[\frac{2}{n^3+1}\]This represents the final simplified result after all cancellable terms are removed.

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Key Concepts

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

Closed Form Expression
When we talk about a **closed form expression**, we're discussing an expression that neatly encapsulates a complex mathematical sequence or series without the need for infinite operations. In essence, it's all about finding a tidy, compact representation of a potentially expansive or complicated problem.
  • Instead of dealing with infinite sums, products, or iterative sequences, a closed-form expression strives to express the result as a concise formula.
  • It provides a quick way to understand and compute the outcome without going through every single term or operation in sequence.
  • In our exercise, the closed form expression is the simplified result of the product after recognizing patterns and performing cancellations.
In this specific exercise, finding \( \frac{2}{n^3+1} \) as the closed form expression means that, regardless of how large \( n \) gets, you can directly calculate the product's outcome with this compact equation. It saves time and provides clarity in both understanding and computation.
Factorization
**Factorization** is one of the fundamental tools in mathematics aimed at breaking down complex expressions into simpler, multiplicative components. By reducing an expression into factors, it often becomes much easier to manipulate or simplify further.
  • In algebra, factorization involves expressing a mathematical object, like a polynomial, as a product of other objects, or factors, that when multiplied together give the original product.
  • For example, the expression \( k^3 - 1 \) can be factorized into \( (k-1)(k^2 + k + 1) \), making it easier to work with in calculations.
  • In the exercise, factorization helps us in identifying terms that can cancel out across the product, which is critical in simplifying the expressions step-by-step.
Through factorization, complex fractions such as \( \frac{k^3-1}{k^3+1} \) are expressed in terms of simpler polynomials, revealing hidden cancellation opportunities in a product expression.
Telescoping Series
In mathematics, a **telescoping series** is a series in which most terms cancel out when added, leaving only a few terms to be summed up at the end. This principle is particularly useful in simplifying very long series or products.
  • The term 'telescoping' is inspired by the collapsing effect seen in a folding telescope.
  • By strategically evaluating terms, particularly noting the relationships between consecutive terms, large sections of the expressions collapse or cancel out.
  • In our problem, once we factorized and reorganized the terms, much of the series cancels itself out, a classic sign of a telescoping series structure.
Seeing the product unfold in a telescoping manner, as with the adjacent cancellation of terms like \( (k^2 + k + 1) \) and \( (k^2 - k + 1) \), simplifies our job to only the first and last few terms surviving the cancellation battle.
Simplification Techniques
**Simplification techniques** in mathematics play a key role in reducing complex expressions to simpler, more manageable forms. They allow mathematicians to more easily interpret, evaluate, or further manipulate expressions.
  • The main goal of simplification is to present an equation or expression in the simplest way possible, often enhancing clarity and understanding.
  • Techniques include canceling out matching numerator and denominator terms in fractions and merging like terms in algebraic expressions.
  • In the problem at hand, after factorization and capitalizing on the telescoping nature of the product, simplification results in a final straightforward expression.
Ultimately, by using simplification techniques involving factorization and telescoping, the originally complex product \( \prod_{k=1}^{n} \frac{k^3-1}{k^3+1} \) is boiled down to the simple form \( \frac{2}{n^3+1} \), making understanding and computation of such products accessible and immediate.

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Most popular questions from this chapter

Obtain a formula for the sum $$ \sum_{k=1}^{\infty} \frac{\alpha r+\beta}{k(k+1)(k+2)} $$

For what values of \(p\) and \(q\) are you able to establish the convergence of the product of the two series $$ \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k+1)^{p}} \text { and } \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k+1)^{q}} ? $$

Obtain a proof that every series \(\sum_{k=1}^{\infty} a_{k}\) for which \(\sum_{k=1}^{\infty}\left|a_{k}\right|\) converges must itself be convergent without using the Cauchy criterion.

This collection of exercises develops some convergence properties of power series; that is, series of the form $$ \sum_{k=0}^{\infty} a_{k} x^{k}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\ldots $$ A full treatment of power series appears in Chapter \(10 .\) (a) Show that if a power series converges absolutely for some value \(x=x_{0}\) then the series converges absolutely for all \(|x| \leq\left|x_{0}\right|\). (b) Show that if a power series converges for some value \(x=x_{0}\) then the series converges absolutely for all \(|x|<\left|x_{0}\right|\). (c) Let $$ R=\sup \left\\{t: \sum_{k=0}^{\infty} a_{k} t^{k} \text { converges }\right\\} $$ Show that the power series \(\sum_{k=0}^{\infty} a_{k} x^{k}\) must converge absolutely for all \(|x|R\). [The number \(R\) is called the radius of convergence of the series. The explanation for the word "radius" (which conjures up images of circles) is that for complex series the set of convergence is a disk.] (d) Give examples of power series with radius of convergence \(0, \infty, 1,2\), and \(\sqrt{2}\). (e) Explain how the radius of convergence of a power series may be computed with the help of the ratio test. (f) Explain how the radius of convergence of a power series may be computed with the help of the root test. (g) Establish the formula $$ R=\frac{1}{\limsup _{k \rightarrow \infty} \sqrt[k]{\left|a_{k}\right|}} $$ for the radius of convergence of the power series \(\sum_{k=0}^{\infty} a_{k} x^{k}\). (h) Give examples of power series \(\sum_{k=0}^{\infty} a_{k} x^{k}\) with radius of convergence \(R\) so that the series converges absolutely at both endpoints of the interval \([-R, R] .\) Give another example so that the series converges at the right-hand endpoint but diverges at the left-hand endpoint of \([-R, R] .\) What other possibilities are there?

For any two series of positive terms write $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ if \(a_{k} / b_{k} \rightarrow 0\) as \(k \rightarrow \infty\) (a) If both series converge, explain why this might be interpreted by saying that \(\sum_{k=1}^{\infty} a_{k}\) is converging faster than \(\sum_{k=1}^{\infty} b_{k}\). (b) If both series diverge, explain why this might be interpreted by saying that \(\sum_{k=1}^{\infty} a_{k}\) is diverging more slowly than \(\sum_{k=1}^{\infty} b_{k}\). (c) For convergent series is there any connection between $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ and $$ \sum_{k=1}^{\infty} a_{k} \leq \sum_{k=1}^{\infty} b_{k} ? $$ (d) For what values of \(p, q\) is $$ \sum_{k=1}^{\infty} \frac{1}{k^{p}} \preceq \sum_{k=1}^{\infty} \frac{1}{k^{q}} ? $$ (e) For what values of \(r, s\) is $$ \sum_{k=1}^{\infty} r^{k} \preceq \sum_{k=1}^{\infty} s^{k} ? $$ (f) Arrange the divergent series $$ \sum_{k=2}^{\infty} \frac{1}{k}, \sum_{k=2}^{\infty} \frac{1}{k \log k}, \sum_{k=2}^{\infty} \frac{1}{k \log (\log k)}, \sum_{k=2}^{\infty} \frac{1}{k \log (\log (\log k))} \ldots $$ into the correct order. (g) Arrange the convergent series $$ \begin{gathered} \sum_{k=2}^{\infty} \frac{1}{k^{p}}, \sum_{k=2}^{\infty} \frac{1}{k(\log k)^{p}}, \sum_{k=2}^{\infty} \frac{1}{k \log k(\log (\log k))^{p}} \\ \sum_{k=2}^{\infty} \frac{1}{k \log k(\log (\log k))(\log (\log (\log k)))^{p}} \ldots \end{gathered} $$ into the correct order. Here \(p>1\). (h) Suppose that \(\sum_{k=1}^{\infty} b_{k}\) is a divergent series of positive numbers. Show that there is a series $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ that also diverges (but more slowly). (i) Suppose that \(\sum_{k=1}^{\infty} a_{k}\) is a convergent series of positive numbers. Show that there is a series $$ \sum_{k=1}^{\infty} a_{k} \preceq \sum_{k=1}^{\infty} b_{k} $$ that also converges (but more slowly). (j) How would you answer this question? Is there a "mother" of all divergent series diverging so slowly that all other divergent series can be proved to be divergent by a comparison test with that series?
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