Square root, expressed as the limit of a sequence
Let \( c \in \mathbb{R} \) be a real. The sequence defined recursively below converges, and it converges to \( \sqrt{c} \).
Let \( x_1 = c \), and let \( x_n \) be defined like so:
Proof by induction (any other typical methods) can be used to show that sequences defined recursively like this do (or don't) converge.
Intuition
The outer \( \frac{1}{2} \) factor can be seen as taking an average of the two inner terms. Inspecting the second term, \( \frac{1}{2}(c + \frac{c}{c}) = \frac{c + 1}{2} \), the third, \( \frac{1}{2}(\frac{c+1}{2} + c \frac{2}{c+1}) \), and so on, one can be convinced that this sequences is increasing starting from the third term. In addition, it can be seen that the terms don't surpass \( \sqrt{c} \). While not a exactly convincing as a proof, you can check if a term did surpass \( \sqrt{c} \), then the following term would be less than that term.