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I found this interesting equality, but I could not find a way to prove it. Any (beautiful) idea?

$$\lim\limits_{n\to \infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n=\sqrt{ab}$$

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5 Answers 5

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Let $$y=\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n$$ $$\implies \log y= n\log \frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}$$ Let $n=\frac{1}{p}$ as $ n \rightarrow \infty, p\rightarrow 0$ $$\implies \log y= \lim_{p \to 0}\frac{\log \frac{a^p+b^p}{2}}{p}$$ Applying L.H rule, we get $$\log y=\lim_{p \to 0} \frac{a^p\log a+b^p\log b}{a^p+b^p}$$ $$\implies \log y=\frac {1}{2}\log ab$$ $$\implies y=\sqrt {ab}$$

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  • $\begingroup$ @Nehorai thanks, btw I wanted to ask the use of \left and \right in formatting, could you tell? $\endgroup$
    – Nikunj
    Feb 27, 2016 at 22:18
  • $\begingroup$ Yes, e.g without: \left( \right) $(\frac{1}{\sqrt{1-2x}})$ and with: $\left(\frac{1}{\sqrt{1-2x}}\right)$ you can see the difference? $\endgroup$
    – 3SAT
    Feb 27, 2016 at 22:20
  • $\begingroup$ @Nehorai Thanks a lot, the brackets have bothered me for quite a while, but not now $\endgroup$
    – Nikunj
    Feb 27, 2016 at 22:23
  • $\begingroup$ it Was AWESOME! Thank you Nikunj. $\endgroup$
    – MR_BD
    Feb 27, 2016 at 22:23
  • $\begingroup$ @Bidgoli You're welcome $\endgroup$
    – Nikunj
    Feb 27, 2016 at 22:24
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As usual when you see the variable in exponent, the strategy is to take logs. Thus if $L$ is the desired limit then \begin{align} \log L &= \log\left\{\lim_{n \to \infty}\left(\frac{a^{1/n} + b^{1/n}}{2}\right)^{n}\right\}\notag\\ &= \lim_{n \to \infty}\log\left(\frac{a^{1/n} + b^{1/n}}{2}\right)^{n}\text{ (via continuity of log)}\notag\\ &= \lim_{n \to \infty}n\log\left(\frac{a^{1/n} + b^{1/n}}{2}\right)\notag\\ &= \lim_{n \to \infty}n\cdot\dfrac{\log\left(1 + \dfrac{a^{1/n} + b^{1/n} - 2}{2}\right)}{\dfrac{a^{1/n} + b^{1/n} - 2}{2}}\cdot\dfrac{a^{1/n} + b^{1/n} - 2}{2}\notag\\ &= \frac{1}{2}\lim_{n \to \infty}n(a^{1/n} - 1) + n(b^{1/n} - 1)\notag\\ &= \frac{1}{2}(\log a + \log b)\notag\\ &= \frac{1}{2}\log ab\notag \end{align} Hence $L = \sqrt{ab}$. Here I have used two fundamental limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{n \to \infty}n(x^{1/n} - 1) = \log x$$

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  • $\begingroup$ @Dr.MV: I believe that sometimes the most voted answer is not the best one. But no worry I will add my vote for the use of squeeze. $\endgroup$
    – Paramanand Singh
    Feb 28, 2016 at 4:09
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    $\begingroup$ Much appreciated! $\endgroup$
    – Mark Viola
    Feb 28, 2016 at 4:09
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It is worthwhile to observe that this question is the special case of showing that the limit of the $p^{\rm th}$ power mean as $p \to 0^+$ equals the geometric mean, for two numbers $a, b > 0$. Define the $p^{\rm th}$ power mean of a sequence of positive numbers $\boldsymbol x = (x_1, \ldots, x_m)$ for a nonzero real $p$ to be $$M_p(\boldsymbol x) = \left(\frac{1}{m} \sum_{i=1}^m x_i^p \right)^{1/p}.$$ Then we have $$\lim_{p \to 0^+} M_p(\boldsymbol x) = M_0(x) = \left(\prod_{i=1}^m x_i \right)^{1/m},$$ the geometric mean. The proof is given in this Wikipedia article.

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    $\begingroup$ (+1) I read about this somewhere, thank you for linking the proof. $\endgroup$
    – Nikunj
    Feb 27, 2016 at 22:21
  • $\begingroup$ Thanks... i Think "p" in first Formula must be changed to m. $\endgroup$
    – MR_BD
    Feb 27, 2016 at 22:22
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HINT:

Use

$$\begin{align} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n&=\left(\frac{e^{\frac{1}{2n}(\log(a)+\log(b))}\left(e^{\frac1{2n}(\log(a)-\log(b))}+e^{-\frac1{2n}(\log(a)-\log(b))}\right)}{2}\right)^n\\\\ &=\sqrt{ab}\cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right) \tag 1 \end{align}$$

and the squeeze theorem.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Starting from $(1)$, we need to evaluate the limit $$\lim_{n\to \infty}\cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right)$$We can use the inequalities $$1\le \cosh(x)\le \frac{1}{1-x^2}$$for $|x|<1$. Then, for $n$ sufficiently large$$1\le \cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right)\le \left(1-\frac{\log^2\left(\sqrt{a/b}\right)}{n^2}\right)^{-n}\le \frac{1}{1-\frac1n \log^2\left(\sqrt{a/b}\right)}$$whereupon applying the squeeze theorem reveals the limit $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n=\sqrt{ab}}$$as was to be shown! Alternatively, Recalling that $\cosh (x)=1+O(x^2)$, then $$\begin{align}\cosh^n\left(\frac{\log\left(\sqrt{a/b}\right)}{n}\right)&=\left(1+O\left(\frac{1}{n^2}\right)\right)^n\\\\&=e^{n\log\left(1+O\left(\frac{1}{n^2}\right)\right)}\\\\&=e^{O(1/n)}\\\\&\to 1\,\,\text{as}\,\,n\to \infty\end{align}$$Finally, we have $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n=\sqrt{ab}}$$as was to be shown!

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    $\begingroup$ +1 for squeeze, I had not seen it before for $\cosh$"function $\endgroup$
    – Paramanand Singh
    Feb 28, 2016 at 4:10
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Even though there are other solutions, I just wanted to give one that is pretty simple. If we could show that this sequence is decreasing (I suspect the devil is in the details on this one) we could just employ the Arithmetic-Geometric mean inequality. We'd find:

$$ \lim_{n \to \infty} \Big(\frac{ a^{1/n} + b^{1/n} }{2}\big)^{n} \geq \lim_{n \to \infty}\sqrt{ab} = \sqrt{ab} $$

Since this is a decreasing sequence bounded below, the limit must the infimum.

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