Sum of Infinite Surds Recently I came across this question. Compute $$\sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}.$$ 
I know how to do it. By equating the above expression to $x$ (i.e., $x = \sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}$) and $x = \sqrt{10+x}$ and so on. But how is this justified? I mean if we take the expression $\sqrt{10+x}$, literally it is equal to $\sqrt{10+\sqrt{10+\sqrt{100 +\dots}}}$ which alters the original expression. Can someone clarify my doubt?
 A: $x=\sqrt{10+\sqrt{10^2+\sqrt{10^4+\cdots}}}$
$\implies \frac x{\sqrt{10}}=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ (This is the well-known Golden Ratio)
so, $\frac x{\sqrt{10}}=\frac{ 1+\sqrt 5}{2}$

Alternatively, $ \frac x{\sqrt{10}}=\sqrt{1+\frac x{\sqrt{10}}}$ 
Squaring we get, $\frac {x^2}{10}=1+\frac x{\sqrt{10}}\implies x^2-\sqrt{10}x-10=0$
$\implies x=\frac{\sqrt 10\pm \sqrt {50}}{2}=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$
or $y=\sqrt{1+y}$ putting $\frac x{\sqrt{10}}=y$
$\implies y^2=1+y\implies y=\frac{1\pm\sqrt5}2\implies x=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$
But $x>0,$ which means $x=\frac{\sqrt{10}(1+\sqrt 5)}{2}$
For the converge, one may check for "Geometric Infinite Surd" here.
A: In this case, it isn't so simple as putting $x=\sqrt{10+x}$--as you've noted, this changes the expression. Now, if you were looking at $\sqrt{10+\sqrt{10+\sqrt{10+\cdots}}}$, that's precisely what you'd do. The reason we can do that, there, is that we're dealing with a sequence defined recursively by $x_1=\sqrt{10}$, $x_{n+1}=\sqrt{10+x_n}$, and then determining the limit of this sequence. That sequence does converge (as it is increasing and bounded above, for example by $10$), so letting $x$ be the limit, we take $n\to\infty$ in the recursion equation $x_{n+1}=\sqrt{10+x_n}$, yielding $x=\sqrt{10+x}$, which we can then use to determine the limit.

Let's consider the sequence given by $y_1=\sqrt{10}$, $y_{n+1}=\sqrt{10+y_n\sqrt{10}}$. Then $$y_2=\sqrt{10+10}=\sqrt{10+\sqrt{100}},$$ $$y_3=\sqrt{10+\sqrt{10}\sqrt{10+\sqrt{100}}}=\sqrt{10+\sqrt{100+10\sqrt{100}}}=\sqrt{10+\sqrt{100+\sqrt{10,000}}},$$ and so on. Now, if this sequence converges (say to $y$), then we take $n\to\infty$ in the equation $y_{n+1}=\sqrt{10+y_n\sqrt{10}}$, yielding $y=\sqrt{10+y\sqrt{10}}$, which we may solve to determine $y$. The trickier part (in this case) is determining an upper bound, so you can prove convergence. From the sounds of it, though, you're supposed to take for granted that it converges, so all you've got to do is the procedure above.
A: Since the OP expressed some doubts about the procedure not the solution, here are some elements of reassurance. Let $a=10$, or, more generally, any positive real number.
In any interpretation of the exercise, the value of $x$, if it exists, should correspond to the limit of a sequence $(x_n)_{n\geqslant0}$ such that $x_{n+1}=\sqrt{a+\sqrt{a}x_n}$, for every $n\geqslant0$. Here is a fact:

Let $(x_n)_{n\geqslant0}$ denote any sequence defined as above. For every $x_0\geqslant-\sqrt{a}$, $x_n\to\ell_a$, where $\ell_a=\frac12(\sqrt5+1)\sqrt{a}$.

Thus, about any reasonable procedure used to define $x$, first, will succeed, and second, will yield $x=\ell_a$. We happy. (When $a=10$,  $\ell_a=\frac{5+\sqrt5}{\sqrt{2}}$.)
To prove the fact stated above, consider the function $u_a$ defined by $u_a(t)=\sqrt{a+\sqrt{a}t}$, for every $t\geqslant-\sqrt{a}$. Then $u_a$ is continuous, increasing, such that $t\lt u_a(t)\lt\ell_a$ for every $-\sqrt{a}\leqslant t\lt\ell_a$, $u_a(\ell_a)=\ell_a$, and $\ell_a\lt u_a(t)\lt t$ for every $t\gt\ell_a$. 
As a consequence, $x_n=\ell_a$ for every $n\geqslant0$ if $x_0=\ell_a$, $(x_n)_{n\geqslant0}$ is increasing and bounded above by $\ell_a$ and converges to $\ell_a$ for every $-\sqrt{a}\leqslant x_0\lt\ell_a$, and $(x_n)_{n\geqslant0}$ is decreasing and bounded below by $\ell_a$ and converges to $\ell_a$ for every $x_0\gt\ell_a$. In every case, $x_n\to\ell_a$.
