$u_{n+1}=\frac{e^{u_n}}{n+1}$ One can prove that for $x\in \mathbb{R}$, the sequence
$$
u_0=x\text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$
converges to $0$ if $x \in ]-\infty,\delta[$ and diverges to $+\infty$ if $x\in ]\delta,+\infty[$ for a fixed $\delta$. I'm trying to find more information on the value $\delta$ (inequalities or expression) and on the specific sequence
$$
u_0=\delta \text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$
Any reference or help are welcome. The only thing I can prove at the moment is $\ln \ln 2 \le \delta \le 1$.
 A: Numerical results:
It seems that $\delta\approx0.3132776395465557$. This was computed using root-finding techniques to find when $u_n=y$ for $y=0.1,1,10$ and $n=10,100,1000$. Code here.
Proofs of bounds:
Disclaimer: Proof that the bounds are tight is not given, but is supported numerically.
Lower bounds can be proven by observing when $t=u_n=u_{n-1}$ occurs. If this occurs, then all future iterations are clearly decreasing. Solving for this gives
$$t=\frac1ne^t\implies t=-W_{-1}\left(-\frac1n\right)$$
where $W_{-1}$ is the real branch of the Lambert W function which gives the largest value for $t$. Working backwards from this point to find $x$ then gives
$$x=\ln\left(1\cdot\ln\left(2\cdot\ln\left(\dots(n-1)\cdot\ln\left(-nW_{-1}\left(-\frac1n\right)\right)\dots\right)\right)\right)$$
$\delta$ is then at least the supremum of this. Note that there are no solutions for $t<3$. The first few values of this lower bound is given by
$$\begin{array}{c|c}n&\delta\ge{}?\\\hline3&0.1013550034887759\\4&0.2751555022435671\\5&0.3044035425578071\\10&0.3132705224120361\\20&0.3132776395448800\\30&0.3132776395465558\end{array}$$
Note that $n=30$ corresponds to the estimated $\delta$ approximation. See also the code above.
Upper bounds can be shown in a similar manner by solving $u_n=u_{n-1}+1\ge3$, which gives
$$u_{n+1}=\frac1{n+1}e^{u_n}=\frac{en}{n+1}\cdot\frac1ne^{u_{n-1}}=\frac{en}{n+1}\cdot u_n\ge u_n+1$$
and is thus diverges to $\infty$. (Again see code above).
It appears to be the case that these bounds are asymptotically tight. Proving so seems to be rather messy, however.
