How is the approximation $\sqrt{k+1} – \sqrt{k}\approx \frac{1}{2\sqrt{k}}$ done? (suppose $k$ is an integer)
Is this a Taylor expansion?
(P.S. I asked this on Physics Stack Exchange because I encountered this in a physics textbook and I think approximations like this are probably only done in physics/engineering instead of mathematics)
Best Answer
This approximation is valid for large $k\gg 1$.
Consider the small quantity $\frac{1}{k}\ll 1$. Then, by the binomial expansion: $$\sqrt{1+\frac{1}{k}}-1=\left(1+\frac{1}{2k}+\mathcal{O}\left(\frac{1}{k}\right)^2\right)-1\approx\frac{1}{2k}$$ Multiplying both sides by $\sqrt{k}$, we have the desired result.