$a_{n+1} = \sqrt{2 a_{n}}$, 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?

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 
$$\lim_{n\to \infty} a_n = 2$$
