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We have a strictly positive sequence $(a_n)_{n\ge1}.$ We know $$\lim_{n\to\infty}\frac{a_{n+1}}{na_{n}}=a>0.$$ What is the value of: $$\lim_{n\to\infty}\left(\frac{(n+1)^2}{\sqrt[n+1]{a_{n+1}}}-\frac{n^2}{\sqrt[n]{a_n}}\right)?$$ I tried using the theorem when we have all sequence terms positive and we know the limit of $a_{n+1}/a_{n}$ is equal to $L,$ then the limit of square root of rank $n$ of $a_{n}$ is that $L.$ I got that $L = na.$ Then I tried to put that $L$ in the main limit we want to calculate and got $2/a,$ but my teacher said the proof is wrong.

Anne Bauval
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Stefan Solomon
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    Please type the relevant information into the question itself rather than including an image, which messes up search engines and people who use screen readers. – Greg Martin Nov 25 '22 at 16:41
  • I am new to this platform and I dont know latex. I will try to put the information with ASCII characters. – Stefan Solomon Nov 25 '22 at 16:43
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    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Nov 25 '22 at 16:53
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    "square root of rank $n$" -- you mean $n$-th root. Also, how do you get $L=n \cdot a$ form that? It does not make sense because $L$ cannot depend on $n$. – Torsten Schoeneberg Nov 25 '22 at 16:55
  • We cant multiply that limit propriety they gave us with n to say limit of a_{n+1}/a_{n} is n×a? – Stefan Solomon Nov 25 '22 at 17:00
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    No, we cannot. E.g. $\lim_{n \to \infty} 1/n =0$ does not imply $\lim_{n \to \infty} 1 \stackrel{??}=n\cdot 0 =0$. You are not just multiplying with $n$ here, you take the $n$ out of the limit, which is almost always a bad and wrong idea. (Note that multiplying both sides of my example with $n$ actually gives $n \cdot \lim_{n \to \infty} 1/n = n \cdot 0 = 0$ which is still true, albeit useless.) – Torsten Schoeneberg Nov 25 '22 at 17:04
  • I was talking about multiplying the propriety they gave us with the limit as n go infinity of n so we will get limit as n go to infinity of a_{n+1} / a_{n} = the limit as n go infinity of ( n × a ) – Stefan Solomon Nov 25 '22 at 17:10
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    And I just told you you cannot do that, just with a simple example. To take yours: $\lim_{n \to \infty} \dfrac{a_{n+1}}{n a_n} = a$ does not imply $\lim_{n \to \infty} \dfrac{a_{n+1}}{ a_n} = n \cdot a$. Sure you can multiply both sides with $n$, but that gives $n \cdot \lim_{n \to \infty} \dfrac{a_{n+1}}{n a_n} = n \cdot a$ for all $n$, which seems useless. You cannot "move the $n$ out of (or into) the limit". – Torsten Schoeneberg Nov 25 '22 at 17:15
  • I understand, thank you. So we need somehow to get a{n+1}/(n×a_{n}) inside the main limit somehow. But I tried and did not get something nice by forcing. – Stefan Solomon Nov 25 '22 at 17:20
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    We can't use that theorem as we do not know $\frac {a_{n+1}}{a_n}$ converges. In fact we know it can not converge, because if it did we would have $\lim \frac {a_{n+1}}{na_n} = \lim \frac 1nL = 0 \ne a = \lim \frac {a_{n+1}}{na_n}$. SO $\frac {a_{n+1}}{a_n}$ does not converge. – fleablood Nov 25 '22 at 17:22
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    Yes I understand. I feel like the sequence is similar to the factorial sequence. I feel like we need to use somehow Traian Lalescu's limit with factorials but I dont know how. – Stefan Solomon Nov 25 '22 at 17:26
  • See a more general result at https://math.stackexchange.com/q/4485844/72031 – Paramanand Singh Nov 29 '23 at 04:48

2 Answers2

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Let $$p_n:=\frac{n^{2n}}{a_n}.$$ We are looking for $$\lim_{n\to\infty}\left(\frac{(n+1)^2}{\sqrt[n+1]{a_{n+1}}}-\frac{n^2}{\sqrt[n]{a_n}}\right)=\lim_{n\to\infty}\left(\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_n}\right).$$ We know that if $\frac{p_{n+1}}{np_n}\to p>0,$ then $\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_n}\to\frac pe.$

Here, $$\frac{p_{n+1}}{np_n}=\frac{na_n}{a_{n+1}}\left(1+\frac1n\right)^{2n+2}\to p:=\frac{e^2}a$$ hence $$\lim_{n\to\infty}\left(\frac{(n+1)^2}{\sqrt[n+1]{a_{n+1}}}-\frac{n^2}{\sqrt[n]{a_n}}\right)=\frac ea.$$

Anne Bauval
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Here is a treatment of the Lalescu types of limits.

Theorem (O. Carja): Suppose $(a_n:n\in\mathbb{N})$ is a sequence of positive numbers such that
(i) $\lim_n\frac{p_n}{n}=p>0$
(ii) $\lim_n\big(\frac{p_{n+1}}{p_n}\big)^n=q\in\overline{\mathbb{R}}_+$.
Then $$L=\lim_n(p_{n+1}-p_n)$$ exists and $L=p\log q$.

This result is not difficult to prove using basic results from Calculus. A simple and short proof can be found here

To solve the problem the OP, define $$p_n:=\frac{n^2}{\sqrt[n]{a_n}}$$ (i) Then $$\frac{p_n}{n}=\frac{n}{\sqrt[n]{a_n}}=\sqrt[n]{\frac{n^n}{a_n}}\xrightarrow{n\rightarrow\infty}\frac{e}{a}$$ for $$\frac{(n+1)^{n+1}}{a_{n+1}}\frac{a_n}{n^n}=\frac{na_n}{a_{n+1}}\Big(\frac{n+1}{n}\Big)^n\frac{n+1}{n}\xrightarrow{n\rightarrow\infty}\frac{e}{a}.$$ (ii) Also, $$\Big(\frac{p_{n+1}}{p_n}\Big)^n=\frac{(n+1)^{2n}}{n^{2n}}\frac{na_n}{a_{n+1}}\frac{\sqrt[n+1]{a_{n+1}}}{n+1}\frac{n+1}{n}\xrightarrow{n\rightarrow\infty}e^2\frac{1}{a}\frac{a}{e}=e. $$

Applying the theorem above yields the desire limit $$\lim_n(p_{n+1}-p_n)=\frac{e}{a}\log e=\frac{e}{a}$$

Mittens
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  • Wasn't aware that there was a specific theorem to deal with such limits. I mostly used the same proof strategy in deal with such limits. An instance is here: https://math.stackexchange.com/q/4485844/72031 – Paramanand Singh Nov 29 '23 at 04:51
  • @ParamanandSingh: I just tried my luck with the problem in your link and Carja's result worked like a charm. – Mittens Nov 29 '23 at 06:19