Show without a calculator : $e^{-\gamma}<\omega$

Hi I think this question is new :

Problem :

Show that :

$$e^{-\gamma}<\omega$$

Where we have the Euler's number and constant and omega constant wich is the value taking at $x=1$ of the Lambert's function .

I recall two interesting fact on The Lambert function :

$$W_{0}(x)=\sum_{n=1}^{\infty}{\frac{(-n)^{n-1}}{n!}}x^{n}$$

And the Euler-Mascheroni constant can be determined via the Harmonic numbers .

Question :

Can we hope to find a proof by hand without calculator as if Euler meet Lambert at the 18 century ?

Thanks !

The Omega constant $\Omega = W_0(1)$ is the unique solution for real $x$ of $xe^x = 1,$ therefore it is strictly less than the unique positive solution $a$ of the equation $p(a) = 1,$ where $p(x) = x + x^2 + x^3/2 + x^4/6$ (a strictly increasing function of $x$ for $x \geqslant 0$). We will show that $p(4/7) > 1,$ whence $a < 4/7,$ whence: $$ \Omega < 4/7. $$ Proof: $$ p\left(\frac47\right) = \frac47 + \frac{16}{49} + \frac{32}{343}\cdot\frac{25}{21} = \frac{44}{49} + \frac{800}{7203} = \frac{44}{49} + \frac19\left(1 - \frac3{7203}\right) > \frac{44}{49} + \frac19\left(1 - \frac4{49}\right) = 1. $$

Because $\Omega = e^{-\Omega},$ the required inequality is equivalent to $e^{-\gamma} < e^{-\Omega},$ whence (as has been pointed out in the comments) it is equivalent to $\Omega < \gamma.$ It is therefore enough to prove that: $$ \gamma > 4/7. $$

The Wikipedia page for Euler's constant, referring to Duane W. DeTemple, "A Quicker Convergence to Euler's Constant", The American Mathematical Monthly 100 (5), pp. 468–470, states that for each positive integer $n,$ \begin{equation} \label{4337532:eq:1}\tag{1} \frac1{24(n+1)^2} < \gamma_n\left(\frac12\right) - \gamma < \frac1{24n^2}, \end{equation} where for all $\alpha > -n,$ $$ \gamma_n(\alpha) = 1 + \frac12 + \frac13 + \cdots + \frac1n - \log_e(n + \alpha). $$ We only need take $n = 2$ in \eqref{4337532:eq:1}, obtaining: $$ \gamma > \frac{143}{96} - \log_e\left(\frac52\right). $$ Temporarily taking for granted this inequality (whose hairy proof is exiled to a separate section below): \begin{equation} \label{4337532:eq:2}\tag{2} \log_e\left(\frac52\right) < \frac{11}{12}, \end{equation} we conclude $$ \gamma > \frac{143}{96} - \frac{11}{12} = \frac{55}{96} > \frac47, $$ whence $\gamma > \Omega$ (as explained above). $\square$

Proof of \eqref{4337532:eq:2}:

\begin{gather*} \log_e\left(\frac52\right) = \log_e\left(\frac{10}4\right) = \log_e\frac{1 + 3/7}{1-3/7} \\ = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{7\cdot7^7} + \frac{3^9}{9\cdot7^9} + \frac{3^{11}}{11\cdot7^{11}} + \cdots\right) \\ < 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{7\cdot7^7}\left( 1 + \frac{3^2}{7^2} + \frac{3^4}{7^4} + \cdots\right)\right) \\ = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{7\cdot7^7}\cdot\frac{49}{40}\right) = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5} + \frac{3^7}{5\cdot7^6\cdot8}\right) \\ = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5}\left(1 + \frac9{56}\right)\right) = 2\left(\frac37 + \frac{3^3}{3\cdot7^3} + \frac{3^5}{5\cdot7^5}\cdot\frac{65}{56}\right) \\ = \frac67\left(1 + \frac3{49} + \frac{3^4}{5\cdot7^4}\cdot\frac{65}{56}\right) = \frac67\left(1 + \frac3{49} + \frac{3^4\cdot13}{7^5\cdot8}\right). \end{gather*} So we have to prove \begin{gather*} 1 + \frac3{49} + \frac{3^4\cdot13}{7^5\cdot8} < \frac{77}{72}, \text{ i.e., }\ \frac3{49} + \frac{3^4\cdot13}{7^5\cdot8} < \frac5{72}, \text{ i.e., }\ \frac{216}{49} + \frac{3^6\cdot13}{7^5} < 5,\\ \text{ i.e., }\ \frac{20}{49} + \frac{3^6\cdot13}{7^5} < 1, \text{ i.e., }\ \frac{3^6\cdot13}{7^5} < \frac{29}{7^2}, \text{ i.e., }\ 729\cdot13 < 343\cdot29, \end{gather*} i.e., $9477 < 9947$ - which is true. $\square$