Help getting the result $\sqrt{n}-\sqrt{n-\frac{1}{4}}\approx\frac{1}{8\sqrt{n-\frac{1}{8}}}$

Question

I have checked "by hand" that $$\sqrt{n}-\sqrt{n-\frac{1}{4}}\approx\frac{1}{8\sqrt{n-\frac{1}{8}}}$$

However, I do not manage to prove the result. Any help would be welcomed!

Answer 1

$$\sqrt{n}-\sqrt{n-\frac{1}{4}} = \frac{\frac{1}{4}}{\sqrt{n}+\sqrt{n-\frac{1}{4}}}=\frac{\frac{1}{4}}{\sqrt{n-\frac{1}{8}+\frac{1}{8}}+\sqrt{n-\frac{1}{8}-\frac{1}{8}}}$$

$$=\frac{1}{4\sqrt{n-\frac{1}{8}}\sqrt{1+\frac{1}{8n-1}}+4\sqrt{n-\frac{1}{8}}\sqrt{1-\frac{1}{8n-1}}}$$

$$=\frac{1}{4\sqrt{n-\frac{1}{8}}\left(\sqrt{1+\frac{1}{8n-1}}+\sqrt{1-\frac{1}{8n-1}}\right)}$$

$$=\frac{1}{4\sqrt{n-\frac{1}{8}}\left(1+\frac{1}{16n-2}+1-\frac{1}{16n-2}+O(n^{-2})\right)}$$

$$=\frac{1}{8\sqrt{n-\frac{1}{8}}\Bigl(1+O(n^{-2})\Bigr)}\approx\frac{1}{8\sqrt{n-\frac{1}{8}}}$$

Answer 2

All the question asks is to show $$ \begin{align} \lim_{n\to\infty}\left(\sqrt{n}-\sqrt{n-\tfrac14}\right)8\sqrt{n-\tfrac18} &=\lim_{n\to\infty}\frac{2\sqrt{n-\frac18}}{\sqrt{n}+\sqrt{n-\tfrac14}}\\ &=\lim_{n\to\infty}\frac{2\sqrt{1-\frac1{8n}}}{1+\sqrt{1-\tfrac1{4n}}}\\[12pt] &=1 \end{align} $$ However, let's see just how close the approximation is: $$ \begin{align} &\sqrt{n}-\sqrt{n-\tfrac14}-\frac1{8\sqrt{n-\frac18}}\\ &=\frac18\left(\frac2{\sqrt{n}+\sqrt{n-\tfrac14}}-\frac1{\sqrt{n-\frac18}}\right)\\ &=\frac{\left(\sqrt{n-\frac18}-\sqrt{n-\frac14}\right)-\left(\sqrt{n}-\sqrt{n-\frac18}\right)}{8\left(\sqrt{n}+\sqrt{n-\frac14}\right)\sqrt{n-\frac18}}\\ &=\frac{\frac1{\sqrt{n-\frac18}+\sqrt{n-\frac14}}-\frac1{\sqrt{n}+\sqrt{n-\frac18}}}{64\left(\sqrt{n}+\sqrt{n-\frac14}\right)\sqrt{n-\frac18}}\\ &=\frac{\sqrt{n}-\sqrt{n-\frac14}}{64\left(\sqrt{n}+\sqrt{n-\frac14}\right)\sqrt{n-\frac18}\left(\sqrt{n-\frac18}+\sqrt{n-\frac14}\right)\left(\sqrt{n}+\sqrt{n-\frac18}\right)}\\ &=\frac1{256\left(\sqrt{n}+\sqrt{n-\frac14}\right)^2\sqrt{n-\frac18}\left(\sqrt{n-\frac18}+\sqrt{n-\frac14}\right)\left(\sqrt{n}+\sqrt{n-\frac18}\right)}\\ &\le\frac1{4096\left(n-\frac14\right)^{5/2}} \end{align} $$ Below, it is shown that the difference is bounded between $\frac1{4096n^{5/2}}$ and $\frac1{4096\left(n-\frac14\right)^{5/2}}$

Answer 3

Recall:

$√a-√b=\dfrac{a-b}{√a+√b}$, $a,b >0$.

Need to show:

$a_n:=(√n +\sqrt{n-1/4}) \approx 2\sqrt{n-1/8}:=b_n;$

Binomial expansion for large $n$:

$a_n=$

$√n + √n(1-1/(4n))^{1/2}=$

$√n+√n(1-(1/2)(1/(4n)) +O(1/n^2))=$

$=2√n -1/(8√n)+O(1/n^{3/2});$

$b_n=2√n\sqrt{1-1/(8n))}=$

$2√n(1-(1/2)(1/(8n))+O(1/n^2))$

$=2√n-(1/(8√n)+O(1/n^{3/2})$;

$a_n-b_n=O(1/n^{3/2})$