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.
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$\begingroup$ Is a proof based on the fact that $\sqrt[n]{2}\leqslant1+\frac1n$ considered as a proof by induction? $\endgroup$– DidNov 10, 2015 at 15:35
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2$\begingroup$ Hint: $2^{\ln(n)+\gamma}$. $\endgroup$– user65203Nov 10, 2015 at 15:38
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$\begingroup$ Actually I am able to prove this result by induction, but not this way, so please post your answer. $\endgroup$– curious_mindNov 10, 2015 at 15:38
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1$\begingroup$ 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 @.) $\endgroup$– DidNov 11, 2015 at 7:17
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$\begingroup$ @YvesDaoust Nonasymptotic tools, per favor... $\endgroup$– DidNov 11, 2015 at 7:18
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1 Answer
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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: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{n} ≤ log_2 (n+1)$
answered Nov 10, 2015 at 15:42
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$\begingroup$ I Came to this question exactly this way. See this : math.stackexchange.com/questions/1366267/… $\endgroup$ Nov 10, 2015 at 15:45
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$\begingroup$ 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. $\endgroup$ Nov 10, 2015 at 15:47
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$\begingroup$ This merely reports a proof to proving the last inequality, right? $\endgroup$– DidNov 11, 2015 at 7:18