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I'm having trouble with this kind of series: $$a_n = \sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})$$

I tried to make something like perfect square to simplify those $7$th root junk which is obviously impossible, and as $n$ tends to infinity L'Hopital's rule also can't solve the problem in this form.

The multiple choice:

  1. $$\lim_{n\to\infty}a_n=\frac{9}{7}$$
  2. $$\lim_{n\to\infty}a_n=+\infty$$
  3. $$a_n\sim\frac{9}{7}n^{-5/14},\quad n\to+\infty$$
Daniel Clarke
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3 Answers3

3

Rewrite it $$\sqrt{n}\left(\sqrt[7]{n+5} - \sqrt[7]{n-4}\right) = \\ n^{\frac{9}{14}}\left((1+\frac{5}{n})^{\frac{1}{7}}-(1-\frac{4}{n})^{\frac{1}{7}}\right) \\ \approx n^{\frac{9}{14}}\left(1+\dfrac{1}{7}\cdot\dfrac{5}{n} - 1 +\dfrac{1}{7}\cdot\dfrac{4}{n} + o(\dfrac{1}{n})\right) \\ \approx \dfrac{9}{7n^{\frac{5}{14}}}$$. Hence the answer is choice $4$-th down the list.

CiaPan
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Wang YeFei
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  • sir how do you do that , the first aproximation , why 7thRoot of (1+5/n) approx 1+ 1/7*5/n ? – Daniel Clarke Feb 06 '22 at 19:14
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    I used the binomial expansion $(1+x)^{\alpha} \approx 1 + \alpha x + \dfrac{\alpha(\alpha - 1)}{2}x^2 + o(\dfrac{1}{n^2})$, where $\alpha = \dfrac{1}{7}, x = \dfrac{5}{n}, \dfrac{4}{n}$. – Wang YeFei Feb 06 '22 at 19:15
  • oh i see that taylor expansion , you just drop the qradratic and high-order terms out, but i have one last question : when i learn taylor formula section in calculus have not seen the term 'o()" , what's this o(1/n) in this example represent and it's different from the comment you write o(1/n2) ? – Daniel Clarke Feb 06 '22 at 19:31
  • You can look it up these notion by typing little o and big O in google or here at MSE posts. – Wang YeFei Feb 06 '22 at 19:36
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Let $b_n=\sqrt[7]{n+5}$ and let $c_n=\sqrt[7]{n-4}$. Then\begin{align}9&=b_n^{\,7}-c_n^{\,7}\\&=(b_n-c_n)\left(b_n^{\,6}+b_n^{\,5}c_n+b_n^{\,4}c_n^{\,2}+\cdots+c_n^{\,6}\right)\\&\geqslant(b_n-c_n)7\sqrt[7]{n-4}^6,\end{align}and therefore$$a_n=\sqrt n(b_n-c_n)\leqslant\frac{9\sqrt n}{7\sqrt[7]{n-4}^6}.$$Can you take it from here?

José Carlos Santos
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  • sir thank you for your answer, and i sti;; don't figure out 2 points from yours , 1: how do you get the third line (7 7th root n-4 sixth power)from the second line (bn6+bn5cn+bn4cn2.....) 2: the final answer seems still cant use Lhopital rule, how do i use this answer to deal with problem options – Daniel Clarke Feb 06 '22 at 19:04
  • @DanielClarke Notice that for all $n\gt 4$, $c_n\lt b_n$ since $1\lt n-4\lt n+5$. Then $b_n^6+b_n^5c_n+\cdots\gt c_n^6+c_n^5c_n+\cdots=7c_n^6$ – FShrike Feb 06 '22 at 19:07
  • Concerning the first doubt, I used the fact that $b_n>\sqrt[7]{n-4}$ and that $c_n=\sqrt[7]{n-4}$. Concerning the second one, note that$$\frac{\sqrt n}{\sqrt[7]{n-4}^6}=\frac{n^{1/2}}{n^{6/7}}\cdot\frac1{\sqrt[7]{1-4/n}^6}$$and that$$\lim_{n\to\infty}\frac1{\sqrt[7]{1-4/n}^6}=1.$$ – José Carlos Santos Feb 06 '22 at 19:08
  • oh i can see , this question way too complicated than i thought , thank you sir – Daniel Clarke Feb 06 '22 at 19:45
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One more, multistorey, without series, way $$\sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})=$$

$$=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-\sqrt[7]{1-\frac{4}{n}}\right) =\\ =\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-1-\sqrt[7]{1-\frac{4}{n}}+1 \right) =\\ =\sqrt{n}\sqrt[7]{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1}{\frac{5}{n}}\cdot \frac{5}{n}-\frac{\sqrt[7]{1-\frac{4}{n}}-1}{\frac{4}{n}}\cdot \frac{4}{n}\right) =\\ =\frac{\sqrt{n}\sqrt[7]{n}}{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1}{\frac{5}{n}}\cdot 5+\frac{\sqrt[7]{1-\frac{4}{n}}-1}{-\frac{4}{n}}\cdot 4\right) \sim \frac{\sqrt{n}\sqrt[7]{n}}{n}\left(\frac{5}{7}+\frac{4}{7}\right) = \\ =\frac{9}{7n^{\frac{5}{14}}}$$

where it was used $\frac{(1+x)^\alpha-1}{x}\to \alpha, x\to 0$.

zkutch
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