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Prove that : $\dfrac{n+1}{2} \leq 2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}$.

I am unable to prove this even by induction and general method. Indeed, when I look at the question $2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}\leq n+1$, asked by me, I have received a hint as a comment to use binomial theorem and showed $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}\frac1{n^k}\geq1+{n\choose 1}\frac1n=2.$$ So, expression becomes $$2 \cdot \sqrt{2} \cdot \sqrt[3]{2} \cdots \sqrt[n]{2} \leq \left(1 +\dfrac{1}{1}\right)\left(1 +\dfrac{1}{2}\right)\left(1 +\dfrac{1}{3}\right)...\left(1 +\dfrac{1}{n}\right)=n+1.$$ I want to solve this problem exactly by same method. So, for this I've to prove that $$\sqrt[n]{2} \leq \dfrac{n+2}{n+1}.$$ How do I do this ?

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curious_mind
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    The inequality simply isn't true. For large $n$ the RHS will be $O(2^{\ln n}) = O(n^c)$ for some $c<1$. – Erick Wong Nov 13 '15 at 07:22
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    It's false with $n=34$ (and above). – 2'5 9'2 Nov 13 '15 at 07:36
  • @alex.jordan, then please prove the main result somehow. – curious_mind Nov 13 '15 at 07:39
  • @alex.jordan But it should hold for large enough $n$, as your argument shows. – Ewan Delanoy Nov 13 '15 at 07:39
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    What main result? And @EwanDelanoy, what argument? I'm just saying it's false, agreeing with Erick Wong. – 2'5 9'2 Nov 13 '15 at 07:40
  • I entered 2^(1+1/2+...+1/34)-(34+1)/2 into google (all the dots are actual numbers when I entered) and indeed a negative result is returned. The result is around $-0.1338$. – cr001 Nov 13 '15 at 07:40
  • @alex.jordan I was referring to an argument you made in a now deleted comment, but you're right, I've just realized it is wrong too. A standard comparison with an integral shows that the LHS behaves like $(n+1)^{\ln(2)}$ and $\ln(2)$ is not large enough ... to conclude, the inequality is false for all except a finite set of $n$. – Ewan Delanoy Nov 13 '15 at 07:45
  • @alex.jordan when you compare this question with the original question it must be true.... http://math.stackexchange.com/questions/1366267/prove-this-inequality-frac-n2-le-frac11-frac12-frac13-fr/1366272#1366272 – curious_mind Nov 13 '15 at 07:47
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    @EwanDelanoy Oh I see. I deleted that comment like 5 seconds after posting it. Erick's $c$ is actually $1/\log_2(e)$, not $\log_2(c)$ as I mistakenly thought for a second. – 2'5 9'2 Nov 13 '15 at 07:48
  • @Hardey Pandya Just enter the 34 case to any calculator by yourself you you know it cannot be true. The truth of your original question does not imply the truth of this question. – cr001 Nov 13 '15 at 07:52
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    What must be true? This inequality is demonstrably false. $\frac{34+1}{2}=17.5$, and $2\cdot\sqrt{2}\cdots\sqrt[34]{2}\approx17.36\ldots$. So what is the "it" that you are saying must be true? And if you want it proved, do you need to post a new separate question? – 2'5 9'2 Nov 13 '15 at 07:52
  • @alex.jordan No please don't be angry, but I was just telling that... $\frac n2 \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+…+\frac1{2^n - 1} \le n$ here, put $k=2^n-1$ then, $n=log_2(k+1)$ – curious_mind Nov 13 '15 at 07:57
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    Oh, I'm not angry at all. I just don't understand what you are after. If you raise $2$ to the left and center parts of this inequality, you get $(k+1)^{1/2}\leq2\cdot\sqrt{2}\cdots\sqrt[k]{2}$. So you are mistaking what the left side becomes. It's not $\frac{k+1}{2}$. – 2'5 9'2 Nov 13 '15 at 08:02
  • Oh, okay.... then can you prove this ? $(k+1)^{1/2}\leq2\cdot\sqrt{2}\cdots\sqrt[k]{2}$ – curious_mind Nov 13 '15 at 08:04
  • Please post in the answer, with the elementary method (I am not a mathematics major after all so don't know calculus...)...I 'll accept your answer, – curious_mind Nov 13 '15 at 08:06

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Considering the rhs, $$A_n=\prod_{i=1}^n 2^{\frac 1 n}$$ $$\log(A_n)=\sum_{i=1}^n \log(2^{\frac 1 n})=\log(2)\sum_{i=1}^n \frac 1 n=H_n\log(2)$$ For large values of $n$, $$H_n=\gamma +\log(n)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ while the logarithm of the lhs would write $$\log(2)+\log(n) +\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and, using these short expansions, you could see that $\left(\log(rhs)-\log(lhs)\right)$ cancels close to $n=32$ and from that point, becomes negative.

So, as comments already showed it, the inequality only holds for a small range of $n$ (up to some finite $n$ as Did properly pointed out).

Claude Leibovici
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    Asymptotic expansions cannot provide information about the rank $n$ at which RHS becomes smaller than LHS. Replacing "up to $n=33$" by "up to some finite $n$", the approach works. The alternative is to use nonasymptotic estimates of the harmonic numbers and of the logarithm (which is not difficult in the present case). – Did Nov 13 '15 at 09:01
  • @Did. I totally agree and I shall edit my answer accordingly. – Claude Leibovici Nov 13 '15 at 09:03
  • I don't understand... the lhs is simply $\frac{n+1}{2}$, so isn't its logarithm $\log(n+1) - \log(2)$? – A.P. Nov 13 '15 at 09:12
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    @A.P. In order to have similar terms I wrote $\log(n+1)=\log(n)+\log(1+\frac 1n)$ and expanded the second logarithm. – Claude Leibovici Nov 13 '15 at 09:22