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
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$
 A: The solution considers $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[8]{n^2}\right) \tag{1}$$ and $$\lim_{n \to \infty} \left( \sqrt[4]{n} - \sqrt[4]{n+1}\right), \tag{2}$$ yet you are asking about $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[\color{red}{4}]{n^2}\right), \tag{3}$$ which is neither of these.
I will ignore your expression and illustrate how $(1)$ is evaluated in the manner suggested.  Let $a_n = \sqrt[8]{n^2+1}$ and $b_n = \sqrt[8]{n^2}$.  Then because $$(a_n - b_n)(a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7) = a_n^8 - b_n^8, \tag{4}$$ it follows that $$\begin{align}
a_n - b_n 
&= \frac{a_n^8 - b_n^8}{a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7} \\
&= \frac{1}{a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7}. \tag{5} \end{align}$$  Since both $a_n \to \infty$ and $b_n \to \infty$ as $n \to \infty$, the denominator of $(5)$ tends to $\infty$ as $n \to \infty$; hence $$\lim_{n \to \infty} a_n - b_n = 0$$ as claimed.  The same argument applies to $(2)$.  It does not apply to $(3)$.  Why?
A: \begin{align}
 \sqrt[8]{n^2+1}-\sqrt[4]{n+1}&=\big(n^2+1)^{1/8}-(n+1)^{2/8} \\
&=\big(n^2+1)^{1/8}-(n^2+2n+1)^{1/8}\\
&=\frac{\big(1+\tfrac{1}{n^2}\big)^{1/8}-1-\Big(\big(1+\tfrac{2}{n}+\tfrac{1}{n^2}\big)^{1/8}-1\Big)}{\tfrac{1}{n^{1/4}}}
\end{align}
The function $f(x)=x^{1/8}$, $x>0$ has a finite derivative at $x=1$. Letting $h=\frac{1}{n^2}$, we obtain that
$$\frac{f(1+h)-f(1)}{h^{1/8}}=h^{7/8}\frac{f(1+h)-f(1)}{h}\xrightarrow{h\rightarrow0}0\cdot f'(1)=0.$$
Similarly, letting $h=\frac{1}{n}$ we obtain
$$\frac{f(x+2h+h^2)-f(1)}{h^{1/4}}=(2h^{3/4}+h^{7/4})\frac{f(1+2h+h^2)-f(1)}{2h+h^2}\xrightarrow{h\rightarrow0}0\cdot f'(1)=0$$
Hence
$$\lim_{n\rightarrow}\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=0$$
A: Consider a different approach as seen by the following.
\begin{align}
\sqrt[m]{n^p + 1} - \sqrt[m]{n^p} &= \sqrt[m]{n^p} \, \left( \sqrt[m]{1 + \frac{1}{n^p}} - 1 \right) \\
&= \sqrt[m]{n^p} \, \left( e^{\frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right)} - 1 \right) \\
&= \sqrt[m]{n^p} \, \left( \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) +  \frac{1}{2 \, m^2} \, \ln^{2}\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^3}\right) \right) \\
&= \sqrt[m]{n^p} \, \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) \, \left( 1 +  \frac{1}{2 \, m} \, \ln\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^2}\right) \right)
\end{align}
which gives the limit as
\begin{align}
\lim_{n \to \infty} \left( \sqrt[m]{n^p + 1} - \sqrt[m]{n^p} \right) &= \lim_{n \to \infty} \, \sqrt[m]{n^p} \, \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) \, \left( 1 +  \frac{1}{2 \, m} \, \ln\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^2}\right) \right) \\
&= \frac{1}{m} \, \ln(1 + 0) \, \left(1 + \frac{\ln(1)}{2 \, m} + \cdots \right) \times \left(\lim_{n \to \infty} \, \sqrt[m]{n^p} \right) \ \\
&= 0.
\end{align}
Now, for the main problem:
\begin{align}
\sqrt[8]{n^2 + 1} - \sqrt[4]{n+1} &= \sqrt[8]{n^2 + 1} - n^{1/4} + n^{1/4} - \sqrt[4]{n+1} \\
&= ( \sqrt[8]{n^2 + 1} - \sqrt[8]{n^2}) - (\sqrt[4]{n+1} - \sqrt[4]{n})
\end{align}
such that
\begin{align}
\lim_{n \to \infty} \left( \sqrt[8]{n^2 + 1} - \sqrt[4]{n+1} \right) &= \lim_{n \to \infty} \left( \sqrt[8]{n^2 + 1} - \sqrt[8]{n^2} \right) - \lim_{n \to \infty} \left( \sqrt[4]{n + 1} - \sqrt[4]{n} \right) \\
&= (0) - (0) = 0.
\end{align}
A: 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.
