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Assume that $x \in\ [1,\infty)$ and that $n \ge 1$.

How to show that $\sqrt[\leftroot{-2}\uproot{2}n]{x} -1$ is asymptotic with $\ln (x) / n$ when $n \to \infty$ ?

And, if possible, how much error is involved in approximating $\sqrt[\leftroot{-2}\uproot{2}n]{x} -1$ by $\ln (x) / n$ when $n$ is large but not necessarily $\infty$.

Ramiro Magno
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  • Limit of $n\to\infty$ of $f(n)$ does not depend on $n$. In what sense do you want it to tend to $\ln(x)/n$? – Thomas Andrews Feb 19 '17 at 17:41
  • @ThomasAndrews: In the sense of asymptotic analysis: https://en.wikipedia.org/wiki/Asymptotic_analysis. – Ramiro Magno Feb 19 '17 at 17:45
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    The words "tends to" is not used in that article. I suggest you use "is asymptotic with" rather than "tends to." For example, you might have meant that $\sqrt[n]{x}-1-\ln(x)/n\to 0$. (which actually is true, too.) – Thomas Andrews Feb 19 '17 at 18:20
  • Thank you for your comment, indeed saying "is asymptotic with" is clearer. – Ramiro Magno Feb 19 '17 at 18:26
  • Note that the equation $\lim_{n\to\infty}n(x^{1/n} - 1) = \log x$ holds and one can develop the entire theory of $\log x$ starting from this equation. Otherwise we can also show using Riemann sum that the limit is equal to $\int_{1}^{x}(dt/t)$. – Paramanand Singh Feb 20 '17 at 01:14

2 Answers2

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Note that

$$\sqrt[n]{x}-1=e^{\frac1n\log(x)}-1\sim \frac{\log(x)}{n}\,\,\text{as}\,\,n\to \infty$$

since $e^x=1+x+O(x^2)$ as $x\to 0$.

In fact, $\sqrt[n]{x}-1$ satisfies the inequalities (SEE THIS ANSWER)

$$\frac{\log(x)}{n}\le \sqrt[n]{x}-1\le \frac{\frac{\log(x)}{n}}{1-\frac{\log(x)}{n}}$$

and hence the error, $\epsilon$, as given by $\epsilon\equiv \sqrt[n]{x}-1-\frac{\log(x)}{n}$ satisfies the bounds

$$0\le \epsilon \le \frac{\log^2(x)}{n(n-\log(x))}$$

Community
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Mark Viola
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Under hypothesis $x\ge 1$ we have $\lim\limits_{n\to\infty}\sqrt[n]{x}=1$, so the limit in question is zero.

szw1710
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