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Let $x>0$, find the limit $$\lim_\limits{n\to{\infty}}{n^2\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)}$$

I use Maclaurin series and find out that the limit is $\ln x$. And this is the answer I get from a math forum:

"Let $x_n=\sqrt[n]{x}-1$, $y_n=\dfrac{1}{n}$, then $(x_n)\to0,(y_n)\to0,(y_n)\downarrow$ when $n\to\infty$.

Use Stolz-Cesaro theorem, we have:

$$\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - 1}}{{\frac{1}{n}}} = \ln x$$ Then $\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[n]{x} - \sqrt[{n + 1}]{x}} \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}}}{{n\left( {n + 1} \right)}}\cdot\dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \ln x$.

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I just wonder what form Stolz-Cesaro theorem that he used, because I just learn Stolz-Cesaro theorem when $(y_n)\to\infty$.

Could anyone help me to get this? Thank you in advance.

This is "another form of Stolz theorem" in the forum that I mentioned: Let $(x_n),(y_n)$ be two sequences of real numbers. Assume $\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } {y_n} = 0$; $y_n>0$, $(y_n)$ is strictly decreasing and the following limit exist $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_{n + 1}} - {x_n}}}{{{y_{n + 1}} - {y_n}}} = L$. Then $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_n}}}{{{y_n}}} = L$. Is this correct?

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  • $\begingroup$ Take a look at my answer at math.stackexchange.com/questions/934774/…, try to use that formula to rationalize your expression. $\endgroup$ – user175968 Feb 28 '17 at 17:28
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    $\begingroup$ Also important: Stolz-Cesaro would only say that if the first limtiexosts, then it equals the second limit $\endgroup$ – Hagen von Eitzen Feb 28 '17 at 17:37
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Indeed, most of us (including me) only learn of the version of the Stolz-Cesàro theorem where $b_n \to \infty$ and $(b_n)_{n \ge 1}$ is strictly increasing. Interestingly, there is another version of this theorem, where $b_n \to 0$ and $(b_n)_{n \ge 1}$ is strictly monotone:

Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be sequences of strictly monotone real numbers with $a_n \to 0$ and $b_n \to 0$. If $\lim \limits _{n \to \infty} \dfrac {a_{n+1} - a_n} {b_{n+1} - b_n} = l \in \bar {\Bbb R}$, then $\left( \dfrac {a_n} {b_n} \right)_{n \ge 1}$ also has a limit and $\lim \limits _{n \to \infty} \dfrac {a_n} {b_n} = l$.

You may find a proof in "Real Analysis on Intervals" by A. D. R. Choudary and C. Niculescu - theorem 2.7.1 (you may even download the relevant chapter from the publisher).

This situation is a perfect parallel to what happens with l'Hospital's theorem, where again there are two versions: one when the denominator tends to $0$, another one when it tends to $\infty$.

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$$\lim_\limits{n\to{\infty}}{n^2\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)}=\lim_{n\to\infty}n/(n+1)\cdot x^{1/(n+1)}\cdot\dfrac{x^{1/n(n+1)}-1}{ 1/n(n+1)}=? $$

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  • $\begingroup$ Not what the OP asks about. $\endgroup$ – Alex M. Feb 28 '17 at 18:20
  • $\begingroup$ +1 for the right approach. Solution given in the question is wrong. $\endgroup$ – Paramanand Singh Mar 1 '17 at 5:31
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You can make use of Taylor expansion:

$$\sqrt[n]{x}=1+\frac{\log(x)}{n}+\frac{\log(x)^2}{2n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sqrt[n+1]{x}=1+\frac{\log(x)}{n}+\frac{\log(x)^2}{2n^2}-\frac{\log(x)}{n^2}+O\left(\frac{1}{n^3}\right)$$ So $$\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)=\frac{\log(x)}{n^2}+O\left(\frac{1}{n^3}\right)$$ Multiplying by $n^2$ and taking the limit finishes the solution.

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The version of Cesaro-Stolz you mention is correct, but it's application for the problem at hand is wrong. The theorem is used to deduce the limit of $a_n/b_n$ from that of $(a_n-a_{n+1})/(b_n-b_{n+1})$ and not the other way round. The right approach to solve the problem is given by user "lab bhattacharjee" in his answer.

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    $\begingroup$ The other way around (reciprocal SC) works if $b_n \to \infty$ and $b_n/b_{n+1} \to L \neq 1$ but that is not the case here. (+1) Also I'm not sure if there is a reciprocal SC when $b_n$ is decreasing. $\endgroup$ – RRL Mar 1 '17 at 5:40
  • $\begingroup$ @RRL: Thanks for the info. I was not aware that a converse of SC existed. +1 for the comment. $\endgroup$ – Paramanand Singh Mar 1 '17 at 6:42
  • $\begingroup$ FYI: I posted a question (that includes a proof). math.stackexchange.com/q/2166597/148510 $\endgroup$ – RRL Mar 1 '17 at 8:07

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