Finding the value to which a sequence converges The question is $f_1=\sqrt2 \ \ \ , \ \ f_{n+1}=\sqrt{2f_n}$, I have to show that it converges to 2. The book proceeds like this:
let $\lim f_n=l$.
We have, $f_{n+1}=\sqrt{2f_n} \implies (f_{n+1})^2=2f_n$.
Also, $\lim f_n=l \implies \lim f_{n+1}=l$.   [HOW ?]
Thus, $l^2=2l \implies l\in [0,2]$.   [???]
Can someone please explain these two steps.
 A: The first part should intuitively be clear as changing as changing the indices of the sequence by $1$ should not change its limit. If you want to be more precise, let $\varepsilon > 0$ be given and try to find an $N \in \mathbb{N}$ such that $\lvert f_{n+1} - l \rvert < \varepsilon$ for $n > N$. This is possible since such an $N$ exists for the sequence $f_n$.
As far as the second claim goes, you probably want that $l \in \{ 0 , 2 \}$, i.e. that $l =0$ or $l = 2$, which follows immediately from the equation $l^2 = 2l$, as these are the only two solutions to that.
A: It is a general trick: if you have a recursive scheme $f_{n+1} = g(f_n)$ and you are asked to show that it converges somewhere, provided that $g$ is continuous you obtain that
$$
  f:=\lim_nf_{n+1} = \lim_n g(f_n) = g(\lim_n f_n) =g(f)
$$
so that $f$ has to be a fixpoint of $g$, that is it has to be a solution of the equation $f = g(f)$. Often it happens that there are several fixed points of $g$, so that you shall realize to which one you do converge based on the initial conditions you are given. This part may be the most tricky, and the method would depend on a specific situation you're in. Note also that $f = g(f)$ is always just a necessary condition for the limit of the sequence (provided it does exist), but is not sufficient.
A: See this.  In short, you may express the limit as
$$2^{1/2 + 1/4 + 1/8 + \cdots} = 2^1 = 2$$
There are very interesting generalizations.
A: Let $g_n = \ln f_n$.
The recurrence becomes
$g_1 = \ln 2/2$,
$g_{n+1} = c+g_n/2$,
where $c = \ln 2/2$.
Calculating the first few terms,
we get
$g_2 = c+g_1/2$,
$g_3 = c + g_2/2
= c+ (c+g_1/2)/2
= 3c/2+g_1/4
$,
$g_4 = c + g_3/2
= c+ (3c/2+g_1/4)/2
= 7c/4+g_1/8
$,
and
$g_5 = c + g_4/2
= c+ (7c/4+g_1/8)/2
= 15c/8+g_1/16
$.
The pattern now seems clear:
$g_n = (2^{n-1}-1)c/2^{n-2}
+ g_1/2^{n-1}
=c(2-1/2^{n-2})
+ g_1/2^{n-1}
$,
and this is easy to prove by induction
from the recurrence.
In the limit,
as $n \to \infty$,
$g_n \to 2c = 2(\ln 2/2) = \ln 2$,
so $f_n \to e^{\ln 2} = 2$.
