We are given this sequence:
$a_1 =1, a_{n+1} = 1 + \frac{1}{a_{n}} \forall \; n\geq 1$
Now before marking this question as "duplicate", consider that I have already referred these sources:
How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$
Convergence of sequences and limits
Now, I am relatively new to Real Analysis and for this particular question, I don't need to determine the limit, only to prove that it is Cauchy and hence convergent.
My approach:
Observe that $a_1 =1, a_2 =2, a_3 = \frac{3}{2}, a_4= \frac{5}{3}$ and so on.
Also observe, $a_{n+1} = 1 + \frac{1}{a_n} \forall n\ge 1$
Multiplying throughout by $a_n$ we get,
$a_{n+1}a_{n} = a_{n} + 1 \; \forall n\ge 1$
Claim: $a_n \; \ge 1 \; \forall \; n\ge 1$
Proof by Induction: Base Case: $n=1:\; a_1 =1 \ge 1$ which is true.
Induction Hypothesis: $a_k \ge 1 \;$ for some $k \in \mathbb{N}, k \ge 1$
Induction Step: To prove: $a_{k+1} \ge 1$ for some $k \in \mathbb{N}, k\ge 1$
Now,
$a_k \ge 1$
$\therefore 0 \le \frac{1}{a_{k}} \le 1$
$\therefore 1 \le 1 + \frac{1}{a_{k}} \le 2$
$\therefore 1 \le a_{k+1} \le 2 $
Therefore by Priniciple of Mathematical Induction,
$a_n \; \ge 1 \; \forall \; n\ge 1$
Now Consider,
$a_{n+1}a_{n} = a_{n} + 1 \ge 1+1 =2 \; \; \forall \; n\ge 1$
Changing indices, we get
$a_{n}a_{n-1} = a_{n-1} + 1 \ge 2 \; \; \forall \; n\ge 2$
Now Consider:
\begin{aligned} &\left|a_{n+1}-a_{n}\right| \\ =&\left|1+\frac{1}{a_{n}}-\left(1+\frac{1}{a_{n-1}}\right)\right| \\ =&\left|\frac{1}{a_{n}}-\frac{1}{a_{n-1}}\right| \\ =&\left|\frac{a_{n}-1-a_{n}}{a_{n} \cdot a_{n-1}}\right| \\ =&\left|\frac{a_{n}-a_{n-1}}{a_{n} \cdot a_{n-1}}\right| \end{aligned}
Since we have that $a_{n}a_{n-1} \ge 2 \; \; \forall \; n\ge 2$
We get, $\left|\frac{a_{n}-a_{n}-1}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \forall n\ge 2$
$\left|a_{n+1}-a_{n}\right| = \left|\frac{a_{n}-a_{n}-1}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \forall n\ge 2$
$\therefore$
$$\left|a_{3}-a_{2}\right| = \left|\frac{a_{2}-a_{1}}{a_{2} \cdot a_{1}}\right| \leqslant \frac{\left|a_{2}-a_{1}\right|}{2} = \frac{1}{2}$$
$\therefore$
$$\left|a_{4}-a_{3}\right| = \left|\frac{a_{3}-a_{2}}{a_{3} \cdot a_{2}}\right| \leqslant \frac{\left|a_{3}-a_{2}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^2} $$
$\therefore$
$$\left|a_{5}-a_{4}\right| = \left|\frac{a_{4}-a_{3}}{a_{4} \cdot a_{3}}\right| \leqslant \frac{\left|a_{4}-a_{3}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^3} $$
$\therefore$
$$\left|a_{n+1}-a_{n}\right| = \left|\frac{a_{n}-a_{n-1}}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^{(n+1)-2}}= \frac{\left|a_{2}-a_{1}\right|}{2^{n-1}} \forall n \ge 2 $$
Hence,
for some $m \ge n$
$$\left|a_m-a_n\right| \leqslant \left|a_{m}- a_{m-1}\right| + \left|a_{m-1}-a_{m-2}\right| + ... \left|a_{n+1}-a_{n}\right| $$ $\;$ $$\leqslant \frac{\left|a_2-a_1\right|}{2^{m-2}} + \frac{\left|a_2-a_1\right|}{2^{m-3}} + ... \frac{\left|a_2-a_1\right|}{2^{n-1}}$$
$$ = \left| a_{2}-a_{1} \right| \cdot \frac{1 - {\frac{1}{2}}^{n-m}}{1 - \frac{1}{2}}$$
Setting $$n = N, m= 2N$$
We get for given $ \epsilon > 0 $
For $N> log_{2}{\frac{2-\epsilon}{\epsilon}}$
We have, $m>n \ge N$
$\left|a_m - a_n\right| < \epsilon $
Hence, this sequence is Cauchy.
Is my approach right?
