Questions: (b) Let (an=fracfn+1fn). Which of the following shows that (an-1=1+frac1an-2) ? (an=fracfn+1fn Rightarrow an-1=fracfnfn-1=fracfn-1+fn-2fn-1=1+fracfn-2fn-1=1+frac1fn-1 / fn-2=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfnfn-1=fracfn+fn-1fn-1=1+fracfnfn-1=1+frac1fn-1 / fn=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfnfn-2=fracfn-1+fn-2fn-2=1+fracfn-1fn-2=1+frac1fn-2 / fn-1=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfn-1fn=fracfn-1fn-1+fn-2=1+fracfn-1fn-2=1+frac1fn-2 / fn-1=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfn+2fn+1=fracfn+1+fnfn+1=1+fracfnfn+1=1+frac1fn / fn+1=1+frac1an-2) Assuming that (leftanright) is convergent, find its limit. (If an answer does not exist, enter DNE.)

(b) Let (an=fracfn+1fn). Which of the following shows that (an-1=1+frac1an-2) ? (an=fracfn+1fn Rightarrow an-1=fracfnfn-1=fracfn-1+fn-2fn-1=1+fracfn-2fn-1=1+frac1fn-1 / fn-2=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfnfn-1=fracfn+fn-1fn-1=1+fracfnfn-1=1+frac1fn-1 / fn=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfnfn-2=fracfn-1+fn-2fn-2=1+fracfn-1fn-2=1+frac1fn-2 / fn-1=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfn-1fn=fracfn-1fn-1+fn-2=1+fracfn-1fn-2=1+frac1fn-2 / fn-1=1+frac1an-2) (an=fracfn+1fn Rightarrow an-1=fracfn+2fn+1=fracfn+1+fnfn+1=1+fracfnfn+1=1+frac1fn / fn+1=1+frac1an-2) Assuming that (leftanright) is convergent, find its limit. (If an answer does not exist, enter DNE.)
Transcript text: (b) Let $a_{n}=\frac{f_{n+1}}{f_{n}}$. Which of the following shows that $a_{n-1}=1+\frac{1}{a_{n-2}}$ ? $a_{n}=\frac{f_{n+1}}{f_{n}} \Rightarrow a_{n-1}=\frac{f_{n}}{f_{n-1}}=\frac{f_{n-1}+f_{n-2}}{f_{n-1}}=1+\frac{f_{n-2}}{f_{n-1}}=1+\frac{1}{f_{n-1} / f_{n-2}}=1+$ $\frac{1}{a_{n-2}}$ $a_{n}=\frac{f_{n+1}}{f_{n}} \Rightarrow a_{n-1}=\frac{f_{n}}{f_{n-1}}=\frac{f_{n}+f_{n-1}}{f_{n-1}}=1+\frac{f_{n}}{f_{n-1}}=1+\frac{1}{f_{n-1} / f_{n}}=1+\frac{1}{a_{n-2}}$ $a_{n}=\frac{f_{n+1}}{f_{n}} \Rightarrow a_{n-1}=\frac{f_{n}}{f_{n-2}}=\frac{f_{n-1}+f_{n-2}}{f_{n-2}}=1+\frac{f_{n-1}}{f_{n-2}}=1+\frac{1}{f_{n-2} / f_{n-1}}=1+$ $\frac{1}{a_{n-2}}$ $a_{n}=\frac{f_{n+1}}{f_{n}} \Rightarrow a_{n-1}=\frac{f_{n-1}}{f_{n}}=\frac{f_{n-1}}{f_{n-1}+f_{n-2}}=1+\frac{f_{n-1}}{f_{n-2}}=1+\frac{1}{f_{n-2} / f_{n-1}}=1+$ $\frac{1}{a_{n}-2}$ $a_{n}=\frac{f_{n+1}}{f_{n}} \Rightarrow a_{n-1}=\frac{f_{n+2}}{f_{n+1}}=\frac{f_{n+1}+f_{n}}{f_{n+1}}=1+\frac{f_{n}}{f_{n+1}}=1+\frac{1}{f_{n} / f_{n+1}}=1+\frac{1}{a_{n-2}}$ Assuming that $\left\{a_{n}\right\}$ is convergent, find its limit. (If an answer does not exist, enter DNE.) $\square$ Need Help? Read It
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Solution

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Solution Steps

To find the limit of the sequence \( \{a_n\} \) given that it is convergent, we start by noting the relationship \( a_{n-1} = 1 + \frac{1}{a_{n-2}} \). If the sequence converges to a limit \( L \), then as \( n \) approaches infinity, \( a_n \), \( a_{n-1} \), and \( a_{n-2} \) all approach \( L \). Therefore, we can set up the equation \( L = 1 + \frac{1}{L} \) and solve for \( L \).

Step 1: Set Up the Equation

Given the relationship \( a_{n-1} = 1 + \frac{1}{a_{n-2}} \), we assume the sequence \( \{a_n\} \) converges to a limit \( L \). Therefore, we set up the equation: \[ L = 1 + \frac{1}{L} \]

Step 2: Solve the Equation

To find the limit \( L \), we solve the equation: \[ L = 1 + \frac{1}{L} \] Multiplying both sides by \( L \) to clear the fraction, we get: \[ L^2 = L + 1 \] Rearranging the equation, we obtain a quadratic equation: \[ L^2 - L - 1 = 0 \]

Step 3: Find the Roots of the Quadratic Equation

Solving the quadratic equation \( L^2 - L - 1 = 0 \) using the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -1 \), we get: \[ L = \frac{1 \pm \sqrt{5}}{2} \]

Step 4: Determine the Convergent Limit

The roots of the quadratic equation are: \[ L = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad L = \frac{1 - \sqrt{5}}{2} \] Since \( \frac{1 - \sqrt{5}}{2} \) is negative and the sequence \( \{a_n\} \) consists of positive terms, the limit must be: \[ L = \frac{1 + \sqrt{5}}{2} \]

Final Answer

The limit of the sequence \( \{a_n\} \) is: \[ \boxed{L = \frac{1 + \sqrt{5}}{2}} \]

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