# $2\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdots\sqrt[n]{2}\leq n+1$

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

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

• Is a proof based on the fact that $\sqrt[n]{2}\leqslant1+\frac1n$ considered as a proof by induction?
– Did
Nov 10, 2015 at 15:35
• Hint: $2^{\ln(n)+\gamma}$.
– user65203
Nov 10, 2015 at 15:38
• Actually I am able to prove this result by induction, but not this way, so please post your answer. Nov 10, 2015 at 15:38
• 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, 2015 at 7:17
• @YvesDaoust Nonasymptotic tools, per favor...
– Did
Nov 11, 2015 at 7:18

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