Trying to parse a Putnam solution from 1995 I’m having trouble parsing the solution for the 1995 Putnam, question A2. The proof proceeds:
The easiest proof uses
``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 +
O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant
times $x^{2}$.) Me: That’s all well and good, and I can arrive at that by the generalized binomial theorem.
The proof continues:
\begin{align*}
\sqrt{x+a}-\sqrt{x} &= x^{1/2}(\sqrt{1+a/x} - 1) \\
&= x^{1/2}(1 + a/2x + O(x^{-2})),
\end{align*}
It would appear that’s an error; it would appear the proof writer dropped the minus 1.
But it continues, hence
$\sqrt{\sqrt{x+a} - \sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))$
Which I’m at a loss how to arrive at.  
 A: OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:
$$\sqrt{x+a}-\sqrt{x}=x^{1/2}(1+\frac{a}{2x}+O(x^{-2}))$$
Now, take the square root:
$$\sqrt{\sqrt{x+a}-\sqrt{x}}=[x^{1/2}(1+\frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Distribute the exponent:
$$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}[(1+\frac{a}{2x}+O(x^{-2}))]^{1/2}$$
Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:
$$(1+z)^{1/2}=1+\frac z 2+O(z^2)\rightarrow \\ (1+\frac a{2x}+O(x^{-2}))^{1/2}=1+\frac{a}{4x}+\frac{1}{2}O(x^{-2})+O(\frac{a^2}{4x^2})=1+\frac{a}{4x}+O(x^{-2})$$
Thus, we get:
$$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}(1+\frac{a}{4x}+O(x^{-2}))$$
Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?
$$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}(1+\frac{a}{4x}+O(x^{-2})-1)$$
$$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}(\frac{a}{4x}+O(x^{-2}))$$
Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!
