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I'm trying to show that:

$$\lim\limits_{n\to\infty}{n(\sqrt[n]{n}-1)} = \infty$$

From what I've tried now, all I end up with is basically rewriting the left term as:

$$\lim\limits_{n\to\infty}{\frac{\sqrt[n]{n}-1}{\frac{1}{n}}}$$

and then applying De L'Hôpital's rule (which gets really messy considering that we're deriving $\sqrt[n]{n}$, since $\frac{d}{dn}\sqrt[n]{n} = -n^{\frac{1}{n}-2}(\ln(n)-1)$).

Is there any "nice and quick" way to solve this?

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4 Answers 4

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Note that $x^{1/x}=e^{(\ln x)/x}$. By looking at the power series expansion of $e^t$, we get that if $x\gt 1$ then $e^{(\ln x)/x}\gt 1+\frac{\ln x}{x}$. Now we are finished.

We can prove $e^t\gt 1+t$ if $t\gt 0$ in other ways, for example by looking at the derivative of $e^t$.

Remark: The advantage of the above approach is not so much quickness, though it is quick. What is useful is that it gives a quite precise idea of the size of $\sqrt[n]{n}-1$ for large $n$.

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You should always check to see if a change in variables would make the problem easier. I will use $n = \frac{1}{h}$.

$$\lim\limits_{n\to\infty}{n(\sqrt[n]{n}-1)} = \lim\limits_{n\to\infty}{n^{1+1/n}-n} = \lim\limits_{h\to0^+}{\frac{1}{h^{1+h}}-\frac{1}{h}}$$

Now just condense the fraction and work as normal.

$$\lim\limits_{h\to0^+}\frac{1}{h\cdot h^h}-\frac{1}{h} = \lim\limits_{h\to 0^+}\frac{1-h^h}{h\cdot h^{h}} $$

L'hopitals rule(s):

$$\lim\limits_{h\to 0^+}\frac{-(1+\log h)}{1+h + h\log h} $$

$$\lim\limits_{h\to 0^+}\frac{-\frac{1}{h}}{2 + \log h} $$

$$\lim\limits_{h\to 0^+}\frac{\frac{1}{h^2}}{\frac{1}{h}} $$

$$\lim\limits_{h\to 0^+}\frac{1}{h} = \infty$$

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  • $\begingroup$ This may not be the quickest way, but I like your solution— it is very easy to follow. +1. $\endgroup$
    – chiru
    Jun 12, 2014 at 3:52
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For large values of $n$, an asymptotic development gives $$\sqrt[n] n=1-\frac{\log \left(\frac{1}{n}\right)}{n}+\frac{\log ^2\left(\frac{1}{n}\right)}{2 n^2}-\frac{\log ^3\left(\frac{1}{n}\right)}{6 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$ from which you can also reach the conclusion.

For sure, you can obtain the same writing, as already told by André Nicolas, that $\sqrt[n] n=e^{\frac{\log(n)}{n}}$ and develop as a Taylor series.

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This is almost obvious if one recognizes that $\lim_{n \to \infty}n(\sqrt[n]{x} - 1) = \log x$. And it is easy to show that the sequence $y_{n} = n(\sqrt[n]{x} - 1)$ is decreasing and hence $y_{n} \geq \log x$. If $z_{n} = n(\sqrt[n]{n} - 1)$ then $z_{n} > y_{n} \geq \log x$ if $n > x$. Since $\log x \to \infty$ as $x \to \infty$, it follows that $z_{n} \to \infty$ as $n \to \infty$.

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