Finding Limit of Nested/Continued Logarithm For a sequence $a_n$ defined by: $$a_1 = \ln(1)$$ $$a_2 = \ln\left(\frac{1}{\ln(2)}+1\right)$$ $$\dots a_n = \ln\left(\frac{1}{\ln(\frac{1}{\ln(\dots 1/\ln(n
))}+1)}+1      \right)$$ with $n$ logarithms.
Could someone offer a solution/hint for the limit: $$\lim_{n\to\infty}a_n$$
Thank you kindly!
Note: My initial attempts involved simplifying to $a_n = ln\left( 1/a_n + 1 \right)$ but I wasn't sure as to where to go from there (not to mention, I noticed this is sort of cheating if we wanted to find the limit purely analytically, since verifying that the limit converges was done through graphing).
Another note: As Mc Cheng already pointed out, the limit does converge to some number, but I also noticed that if we were to subsitute the $n$ in the sequence with another variable $x$, then the limit of increasing values of $a_n$ as $x \to \infty$ seem to flip-flop between being an under-approximation and over-approximation (e.g., $\lim_{x\to\infty} a_4 < \lim_{n\to\infty}a_n,$ but $\lim_{x\to\infty} a_5 > \lim_{n\to\infty} a_n$), and the approximations (obviously) get closer to the actual value of $\lim_{n\to\infty} a_n$, so I wonder if the sequence can be turned into an alternating series?
 A: From your comment you stated the following
$$
a_{n+1} = \ln \left( \frac{1}{a_n} + 1 \right)
$$
Then denote the limit of $a_n$ to be $\gamma$ when $n \to \infty$. By definition it follows that $\lim_{n\to \infty} a_n = \lim_{n \to \infty} a_{n+1}$ hence.
$$
\gamma = \ln\left(\frac{1}{\gamma} + 1\right)
$$
Which gives 
$$
\gamma_1 = 0.80646599423632680877...
$$
$$
\gamma_2 = -1.3499764854011254426...
$$
But $\gamma_2$ is not valid, hence $\gamma_1$ is the limit. 
A: As given in Olba12's answer, there is no closed form for the solution of equation  $$\gamma = \ln\left(\frac{1}{\gamma} + 1\right)$$ (I checked using inverse symbolic calculators) and numerical methods are required.
Considering $$f(\gamma)=\gamma - \ln\left(\frac{1}{\gamma} + 1\right)$$ $$f'(\gamma)=1+\frac{1}{\gamma (\gamma +1)}$$ and using (as the simplest) Newton method with $\gamma_0=1$, the following iterates are generated 
$$\left(
\begin{array}{cc}
 n & \gamma_n \\
 1 & 0.795431453706630 \\
 2 & 0.806420981004432 \\
 3 & 0.806465993496711 \\
 4 & 0.806465994236327  
\end{array}
\right)$$
