Question about solution to Putnam 1995 A2. To provide context Putnam 1995 A2 states:

For what pairs of integers $(a,b)$ of positive real numbers does the improper integral
$$ \int_{b}^{\infty} \left( \sqrt{ \sqrt{x+a \vphantom{b}} -\sqrt{x}} - \sqrt{\sqrt{x} - \sqrt{x-b}} \right) $$
  converge?

The solution then proceeds to make the following claims:
$$(1+x)^{\frac{1}{2}} = 1 + x/2 + O(x^2)$$
So:
$$\sqrt{x-a} - \sqrt{x} = x^{1/2}(\sqrt{1 + a/x} - 1)$$
Subbing in the expansion we have:
$$x^{1/2}(1 + a/(2x) + O(x^{-2}))$$ 
Here I didn't agree, I thought it would be:
$$x^{1/2}(1 + a/(2x) + O(x^{-2}) - 1)$$
which becomes:
$$x^{1/2}(a/(2x) + O(x^{-2}))$$ 
Furthermore, after taking another square root they get:
$$x^{1/4}\left(a/(4x) + O(x^{-2})\right)$$
I would imagine it should be:
$$x^{1/4}\sqrt{a/(2x) + O(x^{-2})}$$
Can someone explain what I'm doing wrong or not understanding?
 A: I do not understand your change in notation; I will stick by the original.  In any case, it is
$$\int_b^{\infty} dx \, \left (\sqrt{\sqrt{x+a}-\sqrt{x}} -\sqrt{\sqrt{x}-\sqrt{x-b}}\right ) $$
Note that we should have $a$ and $b \gt 0$ so that we have real quantities throughout the integration region.
We need to worry about the convergence at $\infty$.  We want to find the conditions under which the iintegrand falls off faster than $1/x$ as $x \to \infty$. 
The way I solved this is to keep using the difference of two squares.  In this respect, everything is exact and we need not appeal to crude approximations until the very end.  For example, we may rewrite the integrand $f(x)$ as
$$\begin{align} f(x) &= \left (\sqrt{\sqrt{x+a}-\sqrt{x}} -\sqrt{\sqrt{x}-\sqrt{x-b}}\right )\\ &= \frac{\sqrt{a}}{\sqrt{\sqrt{x+a}+\sqrt{x}}} - \frac{\sqrt{b}}{\sqrt{\sqrt{x}+\sqrt{x-b}}}  \\ &= \frac{\sqrt{a}\sqrt{\sqrt{x}+\sqrt{x-b}} - \sqrt{b} \sqrt{\sqrt{x+a}+\sqrt{x}} }{\sqrt{\left (\sqrt{x+a}+\sqrt{x} \right )\left (\sqrt{x}+\sqrt{x-b} \right )}} \\ &= \frac{(a-b)\sqrt{x} + a \sqrt{x-b}-b \sqrt{x+a}}{\left (\sqrt{a}\sqrt{\sqrt{x}+\sqrt{x-b}}+\sqrt{b} \sqrt{\sqrt{x+a}+\sqrt{x}}  \right ) \sqrt{\left (\sqrt{x+a}+\sqrt{x} \right )\left (\sqrt{x}+\sqrt{x-b} \right )}}\end{align} $$
Now, let's examine this integrand.  Note that
$$a \sqrt{x-b}-b \sqrt{x+a} = \frac{(a^2-b^2) x - ab (a+b)}{a \sqrt{x-b}+b \sqrt{x+a}} $$
No matter what we do, it seems that the best we have is that the numerator behaves as $O(x)$ and the denominator as $O(x^{5/4})$ as $x \to \infty$, which leads to a nonconvergent integral.  Nevertheless, if $a=b$, then the pieces dependent on $x$ in the numerator vanish, and the denominator still behaves as $O(x^{5/4})$ as $x \to \infty$.  Thus, when $a=b$ we have a convergent integral.
A: As you point out, the solution there is incorrect but it can be salvaged.  It's correct that
$$\sqrt{x+a}-\sqrt{x} = x^{1/2}(a/2x + O(x^{-2}))$$
and so if you factor out the $a/2x$ term you get
$$\sqrt{x+a}-\sqrt{x} = a/2\cdot x^{-1/2}(1 + O(x^{-1})).$$
This means that
$$\sqrt{\sqrt{x+a}-\sqrt{x}} = \sqrt{a/2}\cdot x^{-1/4}(1+O(x^{-1}))$$
where we again used binomial expansion.  You can do the same thing to obtain
$$\sqrt{\sqrt{x}-\sqrt{x-b}} = \sqrt{b/2}\cdot x^{-1/4}(1+O(x^{-1})).
$$
So ultimately the integral becomes
$$\int_b^\infty (\sqrt{a/2}-\sqrt{b/2})x^{-1/4}\,dx + \int_b^\infty O(x^{-5/4})\,dx
.$$
The second integral converges because $5/4 > 1$, and since $0 < 1/4 \le 1$ the first integral will converge precisely when $\sqrt{a/2}-\sqrt{b/2}=0$ or equivalently when $a=b$.
