simplifying $ \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2}=1 $ [duplicate]

Question

$$ \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2}=1 $$

I came across this on a practice standardized test. The question was to evaluate the left hand side, and it turns out the answer was 1. I started to do out the algebra (multiply to get a common denominator, multiply out the numerator, and so on), but it took me a solid 5 minutes to end up with a mess that I couldn't get to simplify down. Each element here is irrational, so the fact that they sum to 1 suggests that there is some relationship between them that I'm just not seeing. Also, the test leaves ~2 minutes per question, so the lengthy algebra seems like its solving this the hard way. I've been looking at this for some time and can't figure out what that relationship is, though.

So I'm posing it here - aside from brute forcing via algebra, is there any simple trick that can show this is equal to 1? Is there some clean relationship that I'm not seeing between these values?

Answer 1

As pointed out in a comment, one always has $$ \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}, $$ just by multiplying the numerator and denominator by $\sqrt{n+1}-\sqrt{n}$ and recognizing the resulting denominator as a difference of two squares. This leads to a simple result, not only for the original series, but for its natural generalization: $$ \sum_{k=K_0}^{K_1}\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sum_{k=K_0}^{K_1}(\sqrt{k+1}-\sqrt{k})=\sqrt{K_1+1}-\sqrt{K_0}. $$ For instance, one immediately has $$ \frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{3+\sqrt{8}}=1 $$ as well.