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Chapter 9: Problem 129

The Fibonacci sequence is defined recursively by \(a_{n+2}=a_{n}+a_{n+1}\), where \(a_{1}=1\) and \(a_{2}=1\). (a) Show that \(\frac{1}{a_{n+1} a_{n+3}}=\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}\). (b) Show that \(\sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}}=1\).

Short Answer

Expert verified
The two proofs are correct based on the recursive rule of the Fibonacci sequence and the property of the telescoping series. The remaining terms in the sequence form a finite value of 0.5.

Step by step solution

01

Part A: Step 1: Applying the Fibonacci Sequence Rule

Let's start with the Fibonacci sequence rule \(a_{n+2}=a_{n}+a_{n+1}\). If we replace \(n\) by \(n+1\), we can get \(a_{n+3}=a_{n+1}+a_{n+2}\).
02

Part A: Step 2: Substituting the Fibonacci Rule into the Equation

Now we replace \(a_{n+3}\) in the equation with \(a_{n+1}+a_{n+2}\). This turns \(\frac{1}{a_{n+1} a_{n+3}}\) into \(\frac{1}{a_{n+1}(a_{n+1} + a_{n+2})}\), which equals to \(\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2}(a_{n+1}+a_{n+2})}\). Finally, put \(a_{n+3}\) back into this expression, we get \(\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}\).
03

Part B: Step 1: Applying Telescoping Series Property

The sequence \(\frac{1}{a_{n+1} a_{n+3}}= \frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}\) forms a telescoping series. This means most of the terms will cancel out when we add them together.
04

Part B: Step 2: Adding the Sequence Together

As a result, when we add the sequence from \(n=0\) to \(\infty\), we have \(\sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}} = \sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+2}} - \frac{1}{a_{n+2} a_{n+3}}\), most terms will cancel each other out. The remaining terms are \(\frac{1}{a_{1} a_{2}} - \frac{1}{a_{2} a_{3}}\) where \(a_{1}=a_{2}=1, a_{3}=2\). So the sum is equal to \(1 - \frac{1}{2} = 0.5\).

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Key Concepts

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

Telescoping Series
A telescoping series is a special type of series where consecutive terms cancel each other out, leaving only the first and last terms in the sequence. This characteristic makes solving such series easier, as it simplifies the computation significantly.
Consider the Fibonacci sequence problem: the expression \( \frac{1}{a_{n+1} a_{n+3}} = \frac{1}{a_{n+1} a_{n+2}} - \frac{1}{a_{n+2} a_{n+3}} \) is a perfect example of a telescoping series.
  • When expanded, many terms in the sequence effectively "telescope" away, cancelling each other out.
  • This leaves behind a neatly reduced expression that shows just the extremes 鈥� the initial and the final terms.

In the provided solution, applying the property of telescoping, we see how most terms across the infinite sequence cancel, resulting in just the first and final sum, simplifying the calculation process to find that the series evaluates to 0.5.
Infinite Series
Infinite series refer to a sum of infinitely many terms. In mathematics, they represent the summation of a sequence where the number of terms is boundless. Not every infinite series converges to a definite number, but finding those that do is a key aspect of series analysis.
Children to the infinite series, telescoping series often bring finite outcomes from ostensibly never-ending terms. The Fibonacci problem illustrates this, as each term reflects the Fibonacci ratio's inherent properties allowing terms to collapse, resulting in a clean sum.
  • In our Fibonacci sequence problem, having an infinite sequence means summation starts at zero and extends indefinitely.
  • The beauty of infinite series lies in the intersection of pattern and simplicity, as visible in the telescoping effect of reducing complex terms to evaluate finite sums, like 1 in our case.

Understanding infinite series provides insights into convergence and divergence, vital in fields such as calculus and real analysis.
Recursive Sequence
A recursive sequence is a sequence of numbers where each term is formulated as a function of preceding terms. The Fibonacci sequence is a classic example of recursion, as each term is defined by summing the two terms before it: \( a_{n+2}=a_{n}+a_{n+1} \).
The beauty of recursion is that it builds complex numerical patterns from simple beginnings. In the Fibonacci sequence:
  • It starts with two initial conditions:\( a_1 = 1 \) and \( a_2 = 1 \).
  • Recursive formulas create a chain-like propagation of values, producing a series where each step depends on the formulation from former values.

Strength in recursion lies in its capability to model growth processes and patterns found in nature and finance. The Fibonacci sequence underlines a recursive process that is straightforward yet generates intricate results that fuel numerous mathematical discussions and applications.

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Most popular questions from this chapter

Consider the sequence \(\left\\{a_{n}\right\\}=\left\\{\frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+(k / n)}\right\\}\). (a) Write the first five terms of \(\left\\{a_{n}\right\\}\). (b) Show that \(\lim _{n \rightarrow \infty} a_{n}=\ln 2\) by interpreting \(a_{n}\) as a Riemann sum of a definite integral.

Given two infinite series \(\sum a_{n}\) and \(\sum b_{n}\) such that \(\sum a_{n}\) converges and \(\Sigma b_{n}\) diverges, prove that \(\Sigma\left(a_{n}+b_{n}\right)\) diverges.

Investigation The interval of convergence of the series \(\sum_{n=0}^{\infty}(3 x)^{n}\) is \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) (a) Find the sum of the series when \(x=\frac{1}{6}\). Use a graphing utility to graph the first six terms of the sequence of partial sums and the horizontal line representing the sum of the series. (b) Repeat part (a) for \(x=-\frac{1}{6}\). (c) Write a short paragraph comparing the rate of convergence of the partial sums with the sum of the series in parts (a) and (b). How do the plots of the partial sums differ as they converge toward the sum of the series? (d) Given any positive real number \(M\), there exists a positive integer \(N\) such that the partial sum \(\sum_{n=0}^{N}\left(3 \cdot \frac{2}{3}\right)^{n}>M\) Use a graphing utility to complete the table. $$ \begin{array}{|l|l|l|l|l|} \hline \boldsymbol{M} & 10 & 100 & 1000 & 10,000 \\ \hline \boldsymbol{N} & & & & \\ \hline \end{array} $$

determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If the interval of convergence for \(\sum_{n=0}^{\infty} a_{n} x^{n}\) is \((-1,1)\), then the interval of convergence for \(\sum_{n}^{\infty} a_{n}(x-1)^{n}\) is \((0,2)\).

Projectile Motion A projectile fired from the ground follows the trajectory given by $$ y=\left(\tan \theta-\frac{g}{k v_{0} \cos \theta}\right) x-\frac{g}{k^{2}} \ln \left(1-\frac{k x}{v_{0} \cos \theta}\right) $$ where \(v_{0}\) is the initial speed, \(\theta\) is the angle of projection, \(g\) is the acceleration due to gravity, and \(k\) is the drag factor caused by air resistance. Using the power series representation $$ \ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots, \quad-1
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