Show $\root{-e}\of{e}<\ln2$ without a calculator

Question

I tried manipulating the expression to come up with inequalities such as $1<e^{\frac{1}{e}}\ln2$.

One idea I have is to show that $\lim_{x\to\infty}\left(\frac{1}{x}+\ln2\right)\left(\frac{1}{x}+1\right)^{\frac{x}{e}}>1$.

$$\lim_{x\to\infty}\frac{\left(1+x\ln2\right)\left(1+x\right)^{\frac{x}{e}}}{x^{1+\frac{x}{e}}}>1$$

I don't know how I could progress.

Answer 1

Beforehand, we will need the following "basic" facts :

• $(A)\;$ $2 < e < 3$;
• $(B)\;$ $e^x$ is a strictly monotonically increasing function;
• $(C)\;$ $x^{-1}$ is a strictly monotonically decreasing function;
• $(D)\,$ $\sqrt{x}$ is a strictly monotonically increasing function over $(1,\infty)$.

First of all, it is to be noted that your inequality is equivalent to the statement $e^{e^{-e/2}} < 2$ thanks to $(B)$.

From these basic facts, we can deduce that $e^{-e/2} < e^{-1} < \frac{1}{2}$, where the first inequality is due to $(A)$ and $(B)$, when the second one comes from $(A)$ and $(C)$. In consequence, one has $e^{e^{-e/2}} < \sqrt{e}$ thanks to $(B)$ again. Now, given that $\sqrt{e} < \sqrt{4}$ by $(A)$ and $(D)$, we can conclude that $e^{e^{-e/2}} < 2$. QED

Answer 2

Euler showed how to approximate log 2 as continued fraction, see Enestrom #606, §23. There we find $\displaystyle log(2) \approx \frac{262}{378} \approx 0.693121$ (without a calculator as requested, lookup only).

$\displaystyle\root{-e}\of{e}$ may be written as $\displaystyle\frac{1}{e^{1/e}}$. The decimal expansion of 1/e is known (lookup w/o calculator), how to approximate $e^x$ by CF is known, to compute this without calculator up to the needed precision may do who likes such exercises.

Result: $$\displaystyle\root{-e}\of{e}\approx 0,6922\lt\log(2)\approx \frac{9}{13}\approx 0.6923$$ confirmed.