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

Here, $n \in \Bbb N$. It can be proven by induction but I want to get this result without use of induction.

Jean-Claude Arbaut
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curious_mind
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  • Is a proof based on the fact that $\sqrt[n]{2}\leqslant1+\frac1n$ considered as a proof by induction? – Did Nov 10 '15 at 15:35
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    Hint: $2^{\ln(n)+\gamma}$. –  Nov 10 '15 at 15:38
  • Actually I am able to prove this result by induction, but not this way, so please post your answer. – curious_mind Nov 10 '15 at 15:38
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    To show that $2\leqslant\left(1+\frac1n\right)^n$, use the binomial theorem $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}\frac1{n^k}\geqslant1+{n\choose 1}\frac1n=2.$$ (Unrelated: Next time, please use @.) – Did Nov 11 '15 at 07:17
  • @YvesDaoust Nonasymptotic tools, per favor... – Did Nov 11 '15 at 07:18
  • @Did, is it okay ? Now, I am telling you that you provided such a beautiful elementary proof. thanks – curious_mind Nov 11 '15 at 18:18
  • @did: why not, this isn't specified ? –  Nov 12 '15 at 07:49
  • @YvesDaoust Yes it is "specified" since the question asks to prove that an inequality holds for every $n\in\mathbb N$, not only for every $n$ large enough. – Did Nov 12 '15 at 09:13
  • Yep, I should have hinted $H_n<\ln(n)+1$. In principle, one could also verify the inequality directly for small $n$ and switch to the asymptotic expression when possible. (I admit that this raises difficult technical issues.) –  Nov 12 '15 at 09:33
  • @Did Can you prove this, without induction ? elementary method is supposed....$ \dfrac{n+1}{2} \leq 2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}\leq n+1$ – curious_mind Nov 13 '15 at 04:00
  • Other problem? Then other question. – Did Nov 13 '15 at 07:06
  • Ok, http://math.stackexchange.com/questions/1526926/dfracn12-leq-2-cdot-sqrt2-cdot-sqrt32-cdot-sqrt42-cdots-sqrtn – curious_mind Nov 13 '15 at 07:09

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Have you already tried this approach?

Rewrite: $2^1*2^{1/2}*2^{1/3}*2^{1/4}*...*2^{1/n}=2^{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{n}}$

Test: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{n} ≤ log_2 (n+1)$

Simpson17866
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  • I Came to this question exactly this way. See this : http://math.stackexchange.com/questions/1366267/prove-this-inequality-frac-n2-le-frac11-frac12-frac13-fr – curious_mind Nov 10 '15 at 15:45
  • But in that, I recieved the proof using calculus , which may be nice but I can't understand that sort of calculus (i mean higher level of calculus), so I thought I can recieve answer for this question on some elementary way. – curious_mind Nov 10 '15 at 15:47
  • This merely reports a proof to proving the last inequality, right? – Did Nov 11 '15 at 07:18