Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $

Question

I struggle for a while solving limit of this chain: $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $

I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $

I made this far: $$ \lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right) =n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$

I will appreciate every help. Thank you

$\sqrt{1+x}=1+{x}/{2}...$ – J. W. Tanner Jan 20 '19 at 04:42 — J. W. Tanner, Jan 20 '19 at 04:42
Does $n→∞$ or some other limit? – Awe Kumar Jha Jan 20 '19 at 04:47 — Awe Kumar Jha, Jan 20 '19 at 04:47

score 2 · Answer 1 · answered Jan 20 '19 at 04:56

Hint

First, let $n=\frac 1x$ and compose Taylor series around $x=0$; back to $n$, get successively $$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}+O\left(\frac{1}{n^4}\right)$$ $$\sqrt{n^2+\sqrt{n^4+1}}=\sqrt{2} n+\frac{1}{4 \sqrt{2} n^3}+O\left(\frac{1}{n^5}\right)$$

score 2 · Accepted Answer · answered Jan 20 '19 at 05:16

$$a_n=\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$ $$=\frac{n^3(\sqrt{n^4+1}-n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$ $$=\frac{n^3}{(\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n)(\sqrt{n^4+1}+n^2)}$$ $$=\frac{1}{(\sqrt{1+\sqrt{1+\frac{1}{n^4}}}+\sqrt{2})(\sqrt{1+\frac{1}{n^4}}+1)}$$ Now, apply $\lim_n\to \infty$ on both sides and get the result.

Hope it helps:)

score 2 · Answer 3 · answered Jan 20 '19 at 05:25

$$ \begin{align}L&=\lim_\limits{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)\\& =\lim_\limits{n\to\infty}n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\\&=\lim_\limits{n\to\infty}\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}\\&=\lim_\limits{n\to\infty}\frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}\\&=\lim_\limits{n\to\infty}\dfrac{-n^5+n^3\sqrt{n^4+1}}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt 2}\\&=\lim_\limits{n\to\infty}\dfrac{-n^5+n^3\cdot n^2\sqrt{1+\dfrac1{n^4}}}{n\sqrt{1+\sqrt{1+\dfrac1{n^4}}}+n\sqrt2}\\&=\lim_\limits{n\to\infty}\dfrac{n^5\left[\sqrt{1+\dfrac1{n^4}}-1\right]}{n\left[\sqrt{1+\sqrt{1+\dfrac1{n^4}}}+\sqrt2\right]}\\&\boxed{\text{Let }n=\dfrac1m,\text{ so as }n\to\infty,m\to 0}\\&=\lim_\limits{m\to0}\dfrac{\sqrt{1+m^4}-1}{m^4\sqrt{1+\sqrt{1+m^4}}+\sqrt2}\\&\boxed{\text{As }x\approx 0,(1+x)^n\approx 1+nx}\\&=\lim_\limits{m\to 0}\dfrac{1+\dfrac12m^4-1}{m^4\sqrt{1+1+\dfrac12m^4}+\sqrt 2}\\&=\lim_\limits{m\to0}\dfrac{\dfrac12}{\sqrt{2+\dfrac12m^4}+\sqrt2}\\&=\lim_\limits{m\to0}\dfrac1{2\sqrt2\sqrt{1+\dfrac14m^4}+2\sqrt2}\\&=\lim_\limits{m\to0}\dfrac1{2\sqrt2\left(1+\dfrac18m^4\right)+2\sqrt2}\\&=\boxed{\dfrac1{4\sqrt2}}\end{align}$$

score 2 · Answer 4 · answered Jan 20 '19 at 05:27

From the numerator and denominator take $-n^{3}$ as common term and then rationalize the numerator again. Once you do that you will be left with :

$$\frac{n^{3}}{(n^{2}+\sqrt{n^{4}+1})(\sqrt{n^{2}+\sqrt{n^{4}+1}}+n\sqrt{2})}$$

Divide the numerator and denominator by $n^{3}$, and once you divide the denominator by $n^{3}$ for the first factor divide n^{2} and the second factor just n. You will get : $$\frac{1}{(1+\sqrt{(1/n^{4})+1})(\sqrt{1+\sqrt{(1/n^{4})+1}}+\sqrt{2})}$$

Apply the limit you will get the answer as $\frac{1}{4\sqrt{2}}$

score 2 · Answer 5 · answered Jan 20 '19 at 05:43

You may proceed as follows transforming the limit into a derivative of a function at $0$.

First set $n = \frac{1}{t}$ and consider $t \to 0^+$:

\begin{eqnarray*} n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) & \stackrel{n=\frac{1}{t}}{=} & \frac{\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^4}+1}}-\sqrt{2}\frac{1}{t}}{t^3}\\ & = & \frac{\sqrt{1+\sqrt{1+t^4}}-\sqrt{2}}{t^4}\\ & \stackrel{y=t^4}{=} & \frac{\sqrt{1+\sqrt{1+y}}-\sqrt{2}}{y}\\ & \stackrel{y\to 0^+}{\longrightarrow} & f'(0) = \left.\frac{1}{4\sqrt{1+\sqrt{1+y}}\cdot \sqrt{1+y}} \right|_{y=0} = \boxed{\frac{1}{4\sqrt{2}}}\mbox{ for } f(y) =\sqrt{1+\sqrt{1+y}}\end{eqnarray*}

score 0 · Answer 6 · answered Jan 20 '19 at 09:54

Like Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $,

$$F=\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4}+1}}-\sqrt{2}n\right)\right) =\lim_{y\to0^+}\dfrac{\sqrt{1+\sqrt{1+y}}-\sqrt2}y$$

Set $\sqrt{1+\sqrt{1+y}}-\sqrt2=u\implies u\to0$

and $1+\sqrt{1+y}=(u+\sqrt2)^2=2+2\sqrt2u+u^2$

$\implies y=(1+2\sqrt2u+u^2)^2-1=4\sqrt2u+O(u^2)$

$$F=\lim_{u\to0^+}\dfrac u{4\sqrt2u+O(u^2)}=?$$

Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $

6 Answers6