Solve $n < e^{6 sqrt{n}}$

Mathematics Asked on November 24, 2020

Find for which values of $n in mathbb{N}$ it holds that $$n < e^{6 sqrt{n}}.$$

I tried to use the inequality $(1 + x) leq e^x$, but from this, I can only find that the inequality holds for $n > 36$. But I need to get $n$ as small as possible.

I also tried the induction on $n$, but I stucked in the induction step. In particular, in showing that $e^{6sqrt{n}} + 1 leq e^{6sqrt{n+1}}$.

I appreciate any help and suggestions.

3 Answers

We have $e^{3x} > 3x ge x$ for $xge 0$. Apply this to $x = sqrt{n}$ to get $e^{3sqrt{n}} > sqrt{n}$. Square both sides and you have $e^{6sqrt{n}} > n$ for all $nge 0$.

Correct answer by jjagmath on November 24, 2020

The inequality holds for all positive $x$, so effectively that a proof seems overkill :-)

(Notice that $0<e^{6sqrt0}$ and $1<dfrac6{2sqrt x}e^{6sqrt x}$.)

Or using Taylor,

$$n<1+6sqrt n+18n+cdots$$

Answered by Yves Daoust on November 24, 2020

(Based on hint from user2661923, see the comments above.)

I set $f(x) = e^{6sqrt{x}} - x$. The first derivation of $f(x)$ is $$f'(x) = 3 frac{e^{6sqrt{x}}}{sqrt{x}} - 1.$$

We note that $f(1) = e^6 - 1 > 0$. Moreover, it is not hard to check that $f'(x) > 0$ for $x in [1, +infty)$. Thus $f$ is increasing on $[1, +infty)$ and $f(1) > 0$. Therefore $f(x) > 0$ for all $x in [1, +infty)$. It implies that $x < e^{6sqrt{x}}.$

And my question easily follows.

Answered by Kapur on November 24, 2020

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