Prove that for any $a,b\in \mathbb{N}$, $\sqrt{a^2+b}=a+\frac{b}{2a+\frac{b}{2a+...}}$. Prove that for any natural numbers $a$ and $b$,   $$\sqrt{a^2+b}=a+\frac{b}{2a+\frac{b}{2a+...}}.$$
This question is taken from Spivak's Calculus Chapter 22 on Infinite sequences, Problem 22-7(c).  Part (b) proves the continued fraction expansion for $\sqrt{2}$.  I cannot seem to generalize.
EDIT:  Some answers are assuming that the limit exists, however proving that the limit exists is the part I really need a solution for.  
 A: Let us set 
$$\tag{1}A:=2a+\frac{b}{\left( 2a+\frac{b}{2a+…}\right)}$$
Thus the equality to be established is $\sqrt{a^2+b}=A-a \ (?)$, i.e., $$\tag{2}A=a+\sqrt{a^2+b} \ \ \ \ (?)$$
But, the "generalized continued fraction" in (1) can be expressed under the form:
$$\tag{3}A=2a+\frac{b}{A}$$
which gives a quadratic equation $A^2-2aA-b=0$ in variable $A$, whose positive solution is (2).

Edit: the convergence of the continued fraction follows from a reasoning on the sequence defined (see (3)) by
$$\tag{4}A_{n+1}:=\varphi(A_n)$$
with $\varphi(x):=2a+\frac{b}{x}$ and any positive $A_0$. 
Let us recall that this fixed point convergence is guaranteed through the classical inequation, consequence of the Mean Value Theorem :
$$|\underbrace{\varphi(A_n)}_{A_{n+1}}-\underbrace{\varphi(A)}_A|< M |A_n-A|$$
where $M=\sup_{x \in V}|\varphi'(x)|$ (V arbitrary small vicinity of limit $A$) under the condition that $M<1$.
Indeed : 
$$\tag{5}|\varphi'(x)|=\frac{b}{x^2} < 1$$
Thus, for $A$ being given by (2), it is sufficient to check that:
$$\tag{6}|\varphi'(A)|=\frac{b}{(a+\sqrt{a^2+b})^2} < 1 \ \ \  \ (?)$$
(the existence of vicinity $V$ of $A$ is guaranteed by the continuity of $\varphi$).
(6) is true, by equivalence with $(a+\sqrt{a^2+b})^2>b$, an inequality that is easily verified (if $a>0$ of course).
A: Denote $X:=2a+\cfrac{b}{2a+\cfrac{b}{2a+\ddots}}$. Then, we have,
$$X-a=a+\frac bX$$
Can you take it from here?
A: You get the continued fraction as limit of $x_{n+1}=a+\frac{b}{a+x_n}$, $x_0=a$, if that limit exists. As rational iterations are not that easy to analyze, linearize them by setting $p_0=a$, $q_0=1$, and separate numerator and denominator in
$$
\frac{p_{n+1}}{q_{n+1}}=a+\frac{b}{a+\frac{p_n}{q_n}}=a+\frac{bq_n}{aq_n+p_n}
=\frac{ap_n+(a^2+b)q_n}{aq_n+p_n}
$$
to get the linear iteration
$$
\begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix}
=
\begin{bmatrix}a&a^2+b\\1&a\end{bmatrix}
\cdot
\begin{bmatrix}p_n\\q_n\end{bmatrix}
$$
with eigenvalues $ a\pm\sqrt{a^2+b}$ and eigenvectors $[\sqrt{a^2+b},\pm 1]^T$.
The resulting sequence of partial fractions is thus
$$
x_n=\sqrt{a^2+b}\cdot\frac{(a+\sqrt{a^2+b})^{n+1}+(a-\sqrt{a^2+b})^{n+1}}{(a+\sqrt{a^2+b})^{n+1}-(a-\sqrt{a^2+b})^{n+1}}
$$
and as $|a+\sqrt{a^2+b}|>|a-\sqrt{a^2+b}|$, this always converges to $\sqrt{a^2+b}$.
A: We can write 
$$\sqrt{a^2 +b} + a = 2a + \sqrt{a^2+b}-a= 2a + \frac{ a^2+b -a^2}{\sqrt{a^2 + b} + a}=2a + \frac{b}{\sqrt{a^2+b}+ a}$$
This may be enough for a proof. 
