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The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

Short Answer

Expert verified
In summary, we found the limit of the ratios of successive terms and alternate terms of the Fibonacci sequence using recurrence relation. We then compared the harmonic series with a Fibonacci-related series and established that the harmonic series is greater than or equal to this Fibonacci-related series. Since the limit of the ratio of alternate Fibonacci terms is greater than zero, we concluded that the harmonic series diverges.

Step by step solution

01

Part a: Find the limit of the ratios of successive terms of the Fibonacci sequence.

To find the limit, we will divide both sides of the recurrence relation by \(f_n\) and take the limit as \(n \rightarrow \infty\). $$\lim_{n\to\infty}\frac{f_{n+1}}{f_n} = \lim_{n\to\infty}\frac{f_{n}+f_{n-1}}{f_n}$$ Let \(\varphi=\lim_{n\to\infty}\frac{f_{n+1}}{f_n}\). We have, $$\varphi = 1 + \lim_{n\to\infty}\frac{f_{n-1}}{f_n} = 1 + \frac{1}{\varphi}$$ Multiplying both sides by \(\varphi\) and rearranging the terms, we get: $$\varphi^2 - \varphi - 1 = 0$$ Now we will solve for \(\varphi\) using the quadratic formula: $$\varphi = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Where \(a=1\), \(b=-1\), and \(c=-1\). Plugging in these values, we get: $$\varphi = \frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)} = \frac{1\pm\sqrt{5}}{2}$$ Since the ratios of Fibonacci terms are positive, we take the positive root, giving us: $$\varphi = \frac{1+\sqrt{5}}{2} \approx 1.618$$
02

Part b: Show the limit of the ratio of alternate Fibonacci terms.

To show this, we start by writing the ratio as a limit: $$\lim_{n\to\infty}\frac{f_{n-1}}{f_{n+1}}$$ From part a, we know that \(\varphi = \lim_{n\to\infty}\frac{f_{n+1}}{f_n}\). Thus, we can write $$\lim_{n\to\infty}\frac{f_{n-1}}{f_{n+1}} = \frac{1}{\lim_{n\to\infty}\frac{f_{n+1}}{f_{n-1}}} = \frac{1}{\varphi^2}$$ Now, using the value of \(\varphi\) from part a, we get: $$\lim_{n\to\infty}\frac{f_{n-1}}{f_{n+1}} = \frac{1}{\frac{1+\sqrt{5}}{2}} = 1- \frac{1}{\varphi}$$ Which is approximately \(0.382\).
03

Part c: Comparing the harmonic series with the Fibonacci-related series.

We can show that the harmonic series is greater than or equal to the Fibonacci-related series by comparing terms: $$\sum_{k=1}^{\infty}\frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty}\frac{f_{k-1}}{f_{k+1}}$$ This inequality holds because the harmonic series groups terms as outlined in the exercise, and each of these groups is greater than or equal to the corresponding term in the Fibonacci-related series.
04

Part d: Using part b to conclude that the harmonic series diverges.

From parts b and c, we know that: $$\sum_{k=1}^{\infty}\frac{1}{k} \geq 1+\sum_{k=1}^{\infty}\frac{f_{k-1}}{f_{k+1}}$$ And $$\lim_{n\to\infty}\frac{f_{n-1}}{f_{n+1}} = 1- \frac{1}{\varphi} \approx 0.382$$ Now, since the limit of the ratio of consecutive terms in the Fibonacci-related series is greater than zero, the series does not converge to zero, and therefore, the harmonic series diverges.

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Key Concepts

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

Recurrence Relation
In mathematics, a recurrence relation is a way of defining the terms in a sequence using the preceding elements. For the Fibonacci sequence, the recurrence relation is expressed as \( f_{n+1} = f_n + f_{n-1} \). This formula tells us that each term is the sum of the two preceding terms. The importance of recurrence relations lies in their ability to simplify complex sequences. By using just the initial terms, here \( f_0 = 1 \) and \( f_1 = 1 \), along with the recurrence rule, you can generate the entire Fibonacci sequence. Recurrence relations are not just limited to the Fibonacci sequence; they can be found in various mathematical contexts, especially in computer algorithms and data structures. They help in understanding how certain processes evolve over time, making them crucial for solving iterative problems.
Limits and Convergence
When dealing with sequences, limits refer to the value that the terms of a sequence approach as the index increases indefinitely. For the Fibonacci sequence, we study the limit of the ratio of successive terms: \( \frac{f_{n+1}}{f_n} \).As \( n \rightarrow \infty \), this ratio approaches the golden ratio \( \varphi \), which is approximately \( 1.618 \). This concept of convergence is fundamental because it shows how the sequence stabilizes around a specific pattern or behavior as it progresses. In general, a sequence is said to converge if its terms get arbitrarily close to a specific value, known as the limit. Convergence is a critical concept because it assures us of the sequence's behavior for large values of \( n \). If a sequence doesn't converge, it may diverge, meaning it grows without bound or behaves erratically.
Harmonic Series
The harmonic series is an infinite series expressed as \( \sum_{k=1}^{\infty} \frac{1}{k} \). This series is fascinating because, despite its simple form, it diverges. This means that as you add more and more terms, the sum grows without limit.The exercise cleverly links the harmonic series with the Fibonacci sequence by showing that grouping terms can develop a comparison with Fibonacci-related terms, such as \( \frac{f_{k-1}}{f_{k+1}} \). While each grouping seems to approach a limit or specific structure, the harmonic series itself defies convergence, illustrating an important principle in series analysis. This behavior of divergence is vital in understanding certain phenomena in mathematics, as it highlights the difference between infinite and finite sums and introduces the intricacies involved in dealing with infinite processes.
Golden Ratio
The golden ratio, denoted by \( \varphi \), emerges naturally in the Fibonacci sequence as the limit of the ratio of successive terms. Its value is \( \frac{1+\sqrt{5}}{2} \), approximately \( 1.618 \). This ratio is considered "golden" due to its frequent occurrence in various fields, from art and architecture to nature and mathematics. The golden ratio is not only aesthetically pleasing but is also significant in solving quadratic equations derived from sequences, such as the Fibonacci sequence's recurrence relation. Its unique properties ensure that it fulfills the equation \( \varphi = 1 + \frac{1}{\varphi} \), as demonstrated in the solution to the exercise. Recognizing and understanding the golden ratio's presence in mathematical sequences allows for a deeper appreciation of the unity and interconnectedness across different mathematical themes and highlights its recurring presence in both natural and human-made structures.

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Most popular questions from this chapter

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)

a. Evaluate the series $$ \sum_{k=1}^{\infty} \frac{3^{k}}{\left(3^{k+1}-1\right)\left(3^{k}-1\right)} $$ b. For what values of \(a\) does the series $$ \sum_{k=1}^{\infty} \frac{a^{k}}{\left(a^{k+1}-1\right)\left(a^{k}-1\right)} $$ converge, and in those cases, what is its value?

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n^{1000}}{2^{n}}\right\\}$$

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