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This question already has an answer here:

Prove that $\ln{n} \lt \sqrt{n}$ for $n \in \mathbb{N}^{*}$.

I have tried to prove that using induction but I really don't know how to do it.

I know how to solve it by creating a function $f(x) = \sqrt{x}-\ln{x}$ and then making the derivative, and then show that is bigger than 0. But I need it solved without math analysis (without derrivative, maybe with induction).

How can I do this? Thank you very much!

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marked as duplicate by rlartiga, tilper, Shailesh, Claude Leibovici, J. M. is a poor mathematician Jan 28 '17 at 9:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ How is $\ln n$ defined outside of math analysis ?! $\endgroup$ – Jack D'Aurizio Jan 27 '17 at 13:07
  • $\begingroup$ Well try using: $e^{\sqrt{n}} = 1+ \frac{\sqrt{n}}{1!} + \frac{n}{2!} + \cdots $ $\endgroup$ – crskhr Jan 27 '17 at 13:09
  • $\begingroup$ Hint: $\ln x = 2 \ln(\sqrt x)$ for $x>0$. $\endgroup$ – TZakrevskiy Jan 27 '17 at 13:12
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Note that $$\ln(x) < \log_2(x).$$ now, it suffices to show that $$\log_2(x) < \sqrt{x} \Leftrightarrow x<2^{\sqrt{x}}$$ substituting $x \mapsto n^2$ we get $$n^2<2^n,$$ which is easily proven by induction for $n \ge 3$. $n = 1, 2$ are trivial.

Update:

As Del pointed out, this answer is wrong. I tried to delete it, but an accepted answer cannot be deleted. It's funny how it is accepted :).

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  • $\begingroup$ For $n \neq 3$. That case can be done separately. $\endgroup$ – Daniel Fischer Jan 27 '17 at 13:32
  • $\begingroup$ @DanielFischer thanks, I fixed the answer. $\endgroup$ – SSepehr Jan 27 '17 at 13:34
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    $\begingroup$ You need $n\le 2^{\sqrt n}$ for every $n$ natural $\endgroup$ – Del Jan 27 '17 at 13:41
  • $\begingroup$ Then if I'm not misunderstanding something you should prove the claim for $n$ equal to the square root of a natural number (if you want to prove the original inequality for every natural number) $\endgroup$ – Del Jan 27 '17 at 14:22
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Check by hand that it's true for $n\le 4$. Then show that $$\ln (n+1)-\ln(n)\le \sqrt{n+1}-\sqrt n$$ for $n\geq 5$, from which the result follows by induction. This is equivalent to $$ \ln\left(1+\frac1n\right)\le \frac{1}{\sqrt n+\sqrt{n+1}}.$$ If you know that $$ \left(1+\frac1n\right)^n\le e$$ then $$\ln\left(1+\frac1n\right)\le \frac1n \le \frac{1}{2\sqrt{n+1}}\leq \frac{1}{\sqrt n+\sqrt{n+1}}$$ because if $n\geq 5$ then $\frac1n \le \frac{1}{2\sqrt{n+1}}$.

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While providing an answer to another question I proved this inequality $$\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1\leq \frac{n-1}{(2n-1)\sqrt{n}+1}<\frac{1}{2\sqrt{n}}$$ or $$\frac{\ln{n}}{2n}\leq \frac{1}{2\sqrt{n}}$$

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