Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus


*

*Verify the following limits

(iii) $\lim\limits_{n\to \infty} \left
 [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$

The solution manual says

$$\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right
 ]$$
$$=\lim\limits_{n\to \infty} \left [\left
 (\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right )+\left
 (\sqrt[4]{n}-\sqrt[4]{n+1}\right )\right ]$$
$$=0+0=0$$
(Each of these two limits can be proved in the same way that
$\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was proved in the
text)

How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right ]=0$?
Note that in the main text, $\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was solved by multiplying and dividing by $(\sqrt{n+1}+\sqrt{n})$ to reach
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$

I want to present you a very general method that does not involve any trick. Basically this is using the taylor expansion of $(1+x)^\alpha$ around $0$ to get en equivalent of the sequence :
\begin{aligned}
(1+x)^{\alpha} &=1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^{2}+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+o\left(x^{n}\right)(x \to 0) \\
\end{aligned}
First let's factorize by the dominant term in each term of the sum :
$$\sqrt[8]{n^2+1}-\sqrt[4]{n+1}= n^{\frac{1}{4}}(1+\frac{1}{n^2})^\frac18 -n^\frac{1}{4} (1+\frac1n)^\frac14  $$
Then we use the Taylor expansion of $(1+x)^\alpha$ around $0$ at order $1$ :
$$ (1+\frac{1}{n^2})^\frac18 = 1+ o(\frac1{n}) $$
$$ (1+\frac1n)^\frac14  = 1+\frac14\frac1n +  o(\frac1{n})  $$
Therefore
\begin{align}
\sqrt[8]{n^2+1}-\sqrt[4]{n+1} &= n^{\frac{2}{8}}(1+\frac{1}{n^2})^\frac18 -n^\frac{1}{4} (1+\frac1n)^\frac14  \\
&= n^{\frac{1}{4}}(1+ o(\frac1{n})) - n^\frac{1}{4}(1+\frac14\frac1n +o(\frac1{n})) \\
&= -\frac14 \frac{1}{n^\frac34}+o(\frac{1}{n^\frac34}) \\
&\sim  -\frac14 \frac{1}{n^\frac34}
\end{align}
Conclusion : $$\boxed{\sqrt[8]{n^2+1}-\sqrt[4]{n+1} \sim -\frac14 \frac{1}{n^\frac34} }$$
Since $ -\frac14 \frac{1}{n^\frac34} \to 0$ then the $\sqrt[8]{n^2+1}-\sqrt[4]{n+1} \to 0$.
And we have a little bonus, we know that it's approaching $0$ from the negative side since  the equivalent is negative. We also know that it converges towards $0$ with a a certain speed given by the equivalent.
