Proving the continued fraction expansion of $\sqrt2$ I am working on a multi-part problem, which shows that $$\sqrt{2} = 1+\frac{1}{2+\frac{1}{2+...}}$$
I have shown that given some rational approximation $\frac{m}{n}$ to $\sqrt2$ , we can take $\frac{m'}{n'} = \frac{m+2n}{m+n}$ and get a better approximation. 
I also know that this iteration goes back and forth between being larger and smaller than $\sqrt2$. Say if we start with $\frac{m}{n}<\sqrt2$, then iterating will give us $$\frac{m}{n}<\frac{m''}{n''} = \frac{3m+4n}{2m+3n}<\sqrt2$$
Finally, I am asked to prove that the sequence obtained by iterating in this fashion, starting with $$1,\frac{3}{2},\frac{7}{5},...$$ which I have shown is given recursively by $$q_1=1, q_{n+1} = 1+\frac{1}{1+q_n} $$
converges to $\sqrt2$.
My professor's "hint" is to consider the "odd" and "even" subsequences, and show that they both converge to the same limit. This makes sense to me, since both sequences are clearly monotone and bounded. Most (all?) of my classmates simply wrote that because they are both monotone and bounded by $\sqrt2$, then they must both converge to $\sqrt2$. However, just because they are bounded above and below, respectively by $\sqrt2$, it does not necessarily mean that $\sqrt2$ is the limit for both (or either!) of them. Some classmates have tried to prove by contradiction that if there were some other limit for one of the subsequences, we could iterate again and get closer to $\sqrt2$. I think that this is a faulty argument, because it assumes that we reach the limit. Remember: the sequence is "fixed" as soon as we choose to start it at $q_1 = 1$ !
My counterargument towards most of my classmates has been the sequence given my $\frac{1}{n}$. The sequence is bounded below by $-99$, and no matter how many times we "iterate" (in this case, just taking the next value of $n$) we can always get closer and closer to $-99$, but that doesn't mean that $-99$ is the limit. 
If someone could offer me a resolution to this problem I would be hugely appreciative. It's been driving me crazy for the past week.
EDIT: I realize this post has been marked as a duplicate, however I think it should be left up, as the answer given is a bit different from that in the "original".
 A: Note that
$$\Bigl|\frac{m'}{n'}-\frac mn\Bigr|
  =\Bigl|\frac{m+2n}{m+n}-\frac mn\Bigr|
  =\Bigl|\frac{m^2-2n^2}{(m+n)n}\Bigr|\ .$$
In the same way
$$\Bigl|\frac{m''}{n''}-\frac{m'}{n'}\Bigr|
  =\Bigl|\frac{(m')^2-2(n')^2}{(m'+n')n'}\Bigr|\ ;$$
substituting back in terms of $m,n$ and doing a bit more algebra gives
$$\Bigl|\frac{m''}{n''}-\frac{m'}{n'}\Bigr|
  =\Bigl|\frac{m^2-2n^2}{(2m+3n)(m+n)}\Bigr|
  <\frac13\Bigl|\frac{m'}{n'}-\frac mn\Bigr|\ .$$
Thus the difference between successive approximations approaches zero; since, as you have already proved, odd and even terms are alternately less than and greater than $\sqrt2$, each must approach $\sqrt2$.
A: $$  
\begin{array}{cccccccccccccccccccccccccccccc}
 & & 1 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & \\
\frac{0}{1} & \frac{1}{0} & & \frac{1}{1} &  &  \frac{3}{2} & &  \frac{7}{5}  & & \frac{17}{12} & &  \frac{41}{29} & &   \frac{99}{70}  & &   \frac{239}{169}  & &   \frac{577}{408}   & &   \frac{1393}{985}  & &   \frac{3363}{2378}   & &   \frac{8119}{5741}  \\
              \\
 & 1 & & -1 &  &  1 & &  -1  & & 1 & &  -1 & &   1  & &   -1  & &   1   & &   -1  & & 1   & &  -1  
\end{array}
$$
Given the convergent $\frac{p}{q},$ what would you think is the $\pm 1$ directly underneath?
