# Interesting technique to find a limit...

• Aug 24th 2010, 01:04 AM
Ares_D1
Interesting technique to find a limit...
Suppose I wanted to find the limit of some sequence, given by $a_{n+1} = \sqrt{2a_n}$. Suppose also that for this particular sequence, we have $a_n \leq a_{n+1} < 2$; i.e., the sequence in increasing and converges to 2. Rearranging the sequence would give $\frac{a_{n+1}^2}{a_n} = 2$. But $\displaystyle \lim _{n \to \infty} a_n = \lim _{n \to \infty} a_{n+1} = L$, and so $\displaystyle \lim _{n \to \infty} (a_n) = \lim _{n \to \infty} \frac{a_{n+1}^2}{a_n} = \frac{L^2}{L} = L = 2$.

I already know that this is the correct answer, but is there anything wrong with my methodology?
• Aug 24th 2010, 01:23 AM
evouga
No. Note, however, two things:

1. You missed a corner case allowing L = 0 as an additional solution.
2. You've only proven that *if* the sequence converges, then it must converge to 2 or 0. You still need to show that it does indeed converge (in this case for any initial value $a_0 >= 0$). In this case that's easy, thanks to some useful inequalities satisfied by the geometric mean.

For example, a similar argument shows that the sequence $a_n = -a_{n-1} + 1$ formally converges to 1/2, but in fact that's only true for $a_0 = 1/2$. The sequence fails to converge for all other initial values.