Spivak, Ch. 22, Problem 7b: Can we assume a sequence has a limit to show what the limit is, or must we first prove that it converges? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus



*In Problem 2-16 we saw that any rational approximation $k/l$ to $\sqrt{2}$ can be replaced by a better approximation
$\frac{k+2l}{k+l}$. In particular, starting with $k=l=1$, we obtain

$$1,\frac{3}{2},\frac{7}{5},\frac{17}{12},...$$
(a) Prove that this sequence is given recursively by
$$a_1=1$$
$$a_{n+1}=1+\frac{1}{1+a_n}$$
(b) Prove that $\lim\limits_{n\to\infty} a_n=\sqrt{2}$. This gives the
so-called continued function expansion
$$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+...}}$$

The solution manual proof of $(b)$ is interesting but a bit long and complicated.
The following is an attempt at the proof. It assumes that $\lim\limits_{n\to\infty} a_{n+1}=l$ and uses the fact that $\lim\limits_{n\to\infty} a_{n+1}=\lim\limits_{n\to\infty} a_{n}$.
$$a_{n+1}=1+\frac{1}{1+a_n}=\frac{2+a_n}{1+a_n}$$
$$\lim\limits_{n\to\infty} a_{n+1}=l=\frac{2+l}{1+l}$$
$$l+l^2=2+l$$
$$l=\sqrt{2}$$
This proof assumes that $\lim\limits_{n\to\infty} a_{n+1}$ exists. This seems fishy. Is it the case that we must prove that ${a_n}$ converges before we can use the above proof?
 A: In order to assert the limit of a sequence, we must know that sequence has a limit in addition to show what is the possible value of the limit.
Here is an example of what can happen when we have not proved a sequence has a limit. Let the sequence be $-1, 1, -1 , 1, \cdots$, i.e., $s_1=1$, $s_{n+1}=-s_n$. Had we assumed the sequence has a limit, we could have let $\lim\limits_{n\to\infty} s_{n+1}=\ell$ and used the fact that $\lim\limits_{n\to\infty} s_{n+1}=\lim\limits_{n\to\infty} s_{n}$, computing
$$\begin{aligned}\lim\limits_{n\to\infty} s_{n+1}&=-\lim\limits_{n\to\infty} s_{n}\\
\ell&=-\ell\\
\ell&=0\end{aligned}$$
It would have been absurd had we claimed that $\lim\limits_{n\to\infty} s_n=0$.
Similarly, for sequence $1,3,7,15,\cdots$, i.e., $t_1=1$, $t_{n+1}=2t_n+1$, we could have computed $\lim\limits_{n\to\infty} t_{n+1}=2\lim\limits_{n\to\infty}t_n+1$, claiming $\lim\limits_{n\to\infty} t_{n}=-1$. Note "the limit" obtained would be $-1$ always even if we set the first term $t_1$ to any other number.

Here is a way to do part (b).
It is obvious that $a_n\ge1$ for all $n$ by induction on $n$.
Consider $d_n=|a_n-\sqrt2|$.
Then $$d_{n+1}=
\left|\frac{2+a_n}{1+a_n}-\sqrt2\right|=\left|\frac{(1-\sqrt2)(a_n-\sqrt2)}{1+a_n}\right|\le \frac12d_n$$
Hence, $0\le d_n\le(1/2)^{n-1}$ by induction on $n$.  So $\lim\limits_{n\to\infty}d_n=0$, i.e., $\lim\limits_{n\to\infty}a_n=\sqrt2$.
