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JH

(a) Show that if $\lim_{n \to \infty} a_{2n} = L$ and $\lim_{n \to\infty} a_{2n + 1} = L,$ then $\{ a_n \}$ is convergent and $\lim_{n \to \infty} a_n = L$.(b) If $a_1 = 1$ and$a_{n + 1} = 1 + \frac {1}{1 + a_n}$find the first eight terms of the sequence $\{ a_n \}$. Then use part (a) to show that $\lim_{n \to \infty} a_n = \sqrt{ 2 }$. This gives the continued fraction expansion$\sqrt{ 2 } = 1 + \frac {1}{ 2 + \frac {1}{2 + \cdot \cdot \cdot}}$

(a) If $n$ is even, $n >n_{0} > 2 \max \left\{n_{1}, n_{2}\right\}+ 1 > 2 \max \left\{n_{1}, n_{2}\right\} \Rightarrow$$\frac{n}{2} > \max \left\{n_{1}, n_{2}\right\} \Rightarrow \frac{n}{2} > n_{1} \Rightarrow\left|a_{2\left(\frac{n}{2}\right)}-L\right| < \epsilon \Rightarrow$$\left|a_{n}-L\right|<\epsilon$If $n$ is odd, $n>n_{0}>2 \max \left\{n_{1}, n_{2}\right\}+1 \Rightarrow n-$$1>2 \max \left\{n_{1}, n_{2}\right\} \Rightarrow \frac{n-1}{2}>\max \left\{n_{1}, n_{2}\right\} \Rightarrow \frac{n-1}{2}>$$n_{2} \Rightarrow\left|a_{2\left(\frac{n-1}{2}\right)+1}-L\right|<\epsilon \Rightarrow\left|a_{n}-L\right|<\epsilon$(b) Consecutive terms approach the same number $L$ as $n \rightarrow \infty,$ so substitute $L$into the formula for $a_{n+1}$ and solve for $L .$

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JH

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Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Boston College

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