How is the sequence 1, 1.4, 1.41, 1.414 generated? In many of the standard textbooks discussing Real Numbers, the Cauchy sequence that converges to $\sqrt{2}$ is given as
1, 1.4, 1.41, 1.414, 1.4142, ...
or
2, 1.5, 1.42, 1.415, 1.4143, ...
My question is how are these sequences generated? In other words, if I have 1, 1.4, 1.41 how do I figure out that the next element of the sequence is 1.414?
 A: The following is an example of a method that you could use to figure this out. The method is called bisection. It is not the fastest, but it is clear how to extract digits from it because it has explicit error bounds.
You know $1^2<2$. You know $2^2>2$. So $1<\sqrt{2}<2$ so its first decimal digit is $1$.
You check $1.5^2>2$. You check $1.25^2<2$. You check $1.375^2<2$. You check $1.4375^2>2$. You check $1.40625^2<2$. Now you know $1.40625<\sqrt{2}<1.4375$ so you know the first two digits are $1.4$.
You continue: you check $1.421875^2>2$. You check $1.4140625^2<2$. You check $1.41796875^2>2$. So $1.4140625<\sqrt{2}<1.41796875$, so you know the first three decimal digits now.
You can keep going; at each time you know $\sqrt{2}$ is in between two numbers getting closer together, so as soon as those numbers have a new digit in common, you know that digit of $\sqrt{2}$. On average it takes $\log_2(10) \approx 3.3$ steps to get a new correct decimal digit. 
At the cost of slightly more iterations, you can make calculations easier (if you're doing it by hand) by rounding the lower bound down and/or the upper bound up. For instance back in the second paragraph of iterations you could have said $1.4<\sqrt{2}<1.44$ and then continued, obtaining $1.41<\sqrt{2}<1.415$ at the end of the third paragraph.
A: Other answers give efficient ways of finding additional terms of the Cauchy sequence, or sequences that may be better than the Cauchy series for practical purposes.
But as far as I'm concerned, the Cauchy sequence has nothing to do with practical computations. It is simply either a way of defining the real numbers or (if you prefer Dedekind cuts as a definition) a way of justifying infinite decimal representations of real numbers.
The Cauchy sequences $1, 1.4, 1.41, 1.414, 1.4142,\ldots$ and
$2, 1.5, 1.42, 1.415, 1.4143, \ldots$ are well defined because after each step you have narrowed down the possible values of $\sqrt2$ to an interval of width $10^{-n}$ for some integer $n,$ and for the next step you say which of the $10$ equal subdivisions of that interval contains $\sqrt2.$
Since the point here is simply that there is an answer, not that we know how to find it quickly, you can appeal to "brute force."
For example, after the upper and lower Cauchy sequences around $\sqrt2$
reach the terms $1.414$ and $1.415,$ respectively, the next term in each sequence has to come from the following set:
$$\{1.4140, 1.4141, 1.4142, 1.4143, 1.4144, 1.4145, 1.4146, 1.4147, 1.4148, 1.4149, 1.4150\}.$$
It takes only a finite number of steps to square all of these values and find out which ones are less than $\sqrt2$ and which are greater.
(You already know two of the squares, so you only have to compute the other nine.)
Once you have done this, you find that
$1.4142^2 < 2 < 1.4143^2,$ so $\sqrt2$ is between 
$1.4142$ and $1.4143,$ which are therefore the next terms in the lower and upper sequences, respectively.
All the rest is optimization of the computations.
A: In general, without more information one cannot produce a term in a sequence using just its previous terms.
One can describe where both of these particular sequences come from, however:
The first comes from truncating the decimal expansion of $\sqrt{2}$ at each successive decimal place. One can give an easy explicit formula for this:
$$a_n := 10^{-n} \lfloor 10^n \sqrt{2} \rfloor .$$
The second is similar, but instead of truncating, i.e., rounding down the nearest number whose decimal expansion has $\leq n$ digits, one rounds up:
$$a_n := 10^{-n} \lceil 10^n \sqrt{2} \rceil .$$
There are infinitely many more Cauchy sequences with limit $\sqrt{2}$. One uses the so-called Babylonian Method, which is itself a specialization of Newton's Method. In this case the terms are defined iteratively, by
$$a_n := \frac{1}{2}\left(a_{n - 1} + \frac{2}{a_{n - 1}}\right) ,$$
where we can take $a_0$ to be any suitable value. Taking the convenient value $a_0 = 2$ gives the sequence
$$2, \frac{3}{2}, \frac{17}{12}, \frac{577}{408}, \ldots ,$$
which converges relatively quickly.
