Answer
See explanation
Work Step by Step
Given ℇ>0,
$\because\lim_{n\rightarrow\infty}{a_{2n}}=L,\ \therefore\exists\ N_1\ s.t.\forall n\geq N_1,\ \left|a_{2n}-L\right|<ℇ.$
$\because\lim_{n\rightarrow\infty}{a_{2n+1}}=L,\ \therefore\exists\ N_2\ s.t.\forall n\geq N_2,\left|a_{2n+1}-L\right|<ℇ.$
Now, for such $N_1,N_2$, let $N=\max{\left(N_1,N_2\right)}$. Then, we have
$\exists\ N\ s.t.\forall n\geq N,\ \left|a_{2n}-L\right|<ℇ.$
$\exists\ N\ s.t.\forall n\geq N,\left|a_{2n+1}-L\right|<ℇ.$
Consider an arbitrary $m\geq N$. Depending on whether m is even or odd:
$If\ m=2k\ for\ some\ k\geq N,then\ \left|a_m-L\right|=\left|a_{2k}-L\right|<ℇ.$
$If\ m=2k+1\ for\ some\ k\geq N,then\ \left|a_m-L\right|=\left|a_{2k+1}-L\right|<ℇ.$
Combine the above description, we obtain:
$\forall\ m\geq N,\ \left|a_m-L\right|<ℇ.$
This directly implies that $\forall\ m\geq N,\left|a_m-L\right|<ℇ.$
Therefore, by the definition of the limit, $\lim_{m\rightarrow\infty}{a_m}=L.$
Thus, we have shown that $\lim_{n\rightarrow\infty}{a_n}=L.$
Part 1: Even Terms
While n=1, $a_2 - a_4 = 18/12-17/12=1/12>0$.
Let $a_{2n} - a_{2n+2} > 0$ established for n = k.
Then, $\because a_{2n} > a_{2n+2} \ \therefore 2 + \frac{1}{1 + a_{2k}} < 2 + \frac{1}{1 + a_{2k+2}} $
$\Rightarrow a_{2(k+1)} - a_{2(k+1)+2} = a_{2k+2} - a_{2k+4} = [1 + \frac{1}{2 + \frac{1}{1 + a_{2k}}}] - [1 + \frac{1}{2 + \frac{1}{1 + a_{2k+2}}}] > 0$
By M.I., $a_{2n} - a_{2n+2} > 0$ establishes, i.e., $\{a_{2n} \}$ is decreasing.
While n =1, $a_2 = \frac{3}{2} > 1$.
Let $a_{2n} > 1$ established for n = k.
Then $a_{2(k+1)} = a_{2k+2} = 1 + \frac{1}{2 + \frac{1}{1 + a_{2k}}} > 1$
By M.I., $\{ a_{2n} \}$ has a lower bound with 1.
Part 2: Odd Terms
While n=1, $a_3 - a_1 = \frac{2}{5} > 0$.
Let $a_{2n+1} - a_{2n-1} > 0$ established for n = k.
Then, $a_{2(k+1)+1} - a_{2(k+1)-1} = a_{2k+3} - a_{2k+1} = [1 + \frac{1}{2 + \frac{1}{1 + a_{2k+1}}}] - [1 + \frac{1}{2 + \frac{1}{1 + a_{2k-1}}}] > 0$
By M.I., $a_{2n+1} - a_{2n-1} > 0$ establishes, i.e., $\{a_{2n - 1} \}$ is increasing.
While n =1, $a_1 = 1 < 2$.
Let $a_{2n-1} < 2$ established for n = k.
Then $a_{2(k+1)-1} = a_{2k+1} = 1 + \frac{1}{2 + \frac{1}{1 + a_{2k-1}}} < 2$
$(Consider \ that \ 0 < 1 - \frac{1}{1 + a_{2k-1}} = \frac{1-\frac{1}{1 + a_{2k-1}}}{2+\frac{1}{1 + a_{2k-1}}} < 1)$
By M.I., $\{ a_{2n-1} \}$ has a upper bound with 2.
Part 3: Find the limit of $\{ a_{2n} \}$ and $\{ a_{2n-1} \}$
By the Monotomic Sequence Theorem, we suppose x belonging to R s.t. $\lim_{n\rightarrow\infty}{a_{2n}=x}$
Then $\lim_{n\rightarrow\infty}{a_{2n+2}=x} = 1 + \frac{1}{2 + \frac{1}{1 + a_{2n}}}$.
$\Rightarrow x = 1 + \frac{1}{2 + \frac{1}{1 + x}}$
$\Rightarrow x = \pm \sqrt{2}$ (Choose the positive one)
Similarly, we do the same thing for $\{ a_{2n-1} \}$ and we will get $\lim_{n\rightarrow\infty}{a_{2n-1}= \sqrt{2}}$
Therefore, by the result of part (a), $\because \lim_{n\rightarrow\infty}{a_{2n} = \lim_{n\rightarrow\infty}{a_{2n-1} = \sqrt{2}}} \ \therefore \lim_{n\rightarrow\infty}{a_n = \sqrt{2}}$.
