$a_{n+1} = \sqrt{2 a_{n}}$, for any choice of $a_{0} > 0$. This sequence is monotone and bounded. And what its limit? Hello guys any help here?
I have this sequence: $a_{n+1} = \sqrt{2 a_{n}}$. This sequence is monotone and bounded for any choice of $a_{0} > 0$. 
And what its limit?
 A: From the recurrence formula, we have:
$$a_1 = 2^{\frac{1}{2}} \cdot a_0^{\frac{1}{2}}$$
$$a_2 = 2^{\frac{1}{2}}\cdot a_1^{\frac{1}{2}} = 2^{\frac{1}{2}} \cdot 2^{\frac{1}{4}} \cdot a_0^{\frac{1}{4}} = 2^{\frac{1}{2}+\frac{1}{4}}\cdot a_0^{\frac{1}{4}}$$
$$a_3 = 2^{\frac{1}{2}} \cdot a_2^{\frac{1}{2}} = 2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}+\frac{1}{8}}\cdot a_0^{\frac{1}{8}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}} \cdot a_{0}^{\frac{1}{8}}$$
So in general
$$a_n = 2^{\frac{1}{2}+\frac{1}{4}+\ldots + \frac{1}{2^n}}\cdot a_0^{\frac{1}{2^n}}$$
Since 
$$\lim_{n\to \infty} \sum_{k=1}^{n}\frac{1}{2^k} = \lim_{n\to \infty} \left(1-\frac{1}{2^{n}}\right) = 1$$
and 
$$\lim_{n \to \infty} a_0^{\frac{1}{2^n}} = a_0^0 = 1$$
we can deduce that 
$$\lim_{n\to \infty} a_n = 2$$
A: Here is another approach. 
Since $a_n$ is monotone and bounded its limit exists.  Call this limit $L$.
Then passing the limit through the recurrence relationship $a_{n+1}=\sqrt{2 a_n}$
we get that $L = \sqrt{2L}$.  Solving for $L$ yields $L^2 - 2L = 0$ which implies 
$L = 0$ or $L=2$.  But $a_0 >0$ so we must have that $L=2$.
A: Similarly, if
$a_{n+1} = (ca_n)^{1/m}
$
then
$a_n \to c^{1/(m-1)}$.
This is $m=2, c=2$.
A: You have $$a_{n+1} = \sqrt{2 a_{n}}\implies a_{n+1}^2=2a_n\implies 2\log(a_{n+1})=\log(2)+\log(a_n)$$ Let $b_n=\log(a_n)$ to get 
$$2b_{n+1}=b_n+\log(2)\implies b_n=C \,2^{1-n}+(1-2^{-n})\log(2)$$ Back to $a_n$
$$a_n=\exp\left(C \,2^{1-n}+(1-2^{-n})\log(2) \right)$$ and then the limit is $e^{\log(2)}=2$
