We know, that $$e=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{1}\left(1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$ If $x_{m}=m(x_{m-1}-1)$ and $x_{0}=e$, then $$x_{1}=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$ $$x_{2}=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$ $$x_{m}=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{m+1}\left(1+\frac{1}{m+2}\left(1+\frac{1}{m+3}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$ $$\sum\limits_{m=0}^{n}\left(x_{m}-1\right)\approx\ln(n)+k$$ How and where can I calculate this constant? Is there relationship with this constant and $\gamma$?
If I made some mistakes, sorry for my English.
For general $m \in \mathbb{N}$, it is easy to see $$x_m = m!\left(e - \sum_{k=0}^{m-1}\frac{1}{k!}\right) \quad\implies\quad x_m - 1 = m!\left(e - \sum_{k=0}^{m} \frac{1}{k!}\right) = \sum_{k=m+1}^\infty \frac{m!}{k!}$$
This leads to $$\begin{align}x_m - 1 = \sum_{k=0}^\infty \frac{m!}{(m+k+1)!} &= \frac{1}{m+1} + \sum_{k=1}^\infty \frac{1}{k!}\frac{\Gamma(m+1)\Gamma(k+1)}{\Gamma(m+k+2)}\\ &= \frac{1}{m+1} + \sum_{k=1}^\infty\int_0^1 (1-t)^m \frac{t^k}{k!} dt\\ &= \frac{1}{m+1} + \int_0^1 (1-t)^m (e^t-1)dt \end{align} $$ Summing $m$ from $0$ to $n$, we get
$$\begin{align} \sum_{m=0}^n (x_m - 1) &= H_{n+1} + \sum_{m=0}^n\int_0^1 (1-t)^m (e^t-1)dt\\ &= H_{n+1} + \int_0^1 \frac{1-(1-t)^{n+1}}{1-(1-t)} (e^t-1) dt\\ &= H_{n+1} + \int_0^1 (1 - (1-t)^{n+1}) \frac{e^t-1}{t}dt \end{align} $$ where $H_n = \sum_{m=1}^n \frac{1}{m}$ is the $n^{th}$ harmonic number.
Notice for $t \in (0,1)$, MVT tell us for some $\xi \in (0,t)$, $\frac{e^t - 1}{t} = e^{\xi t} \in (1, e^t) \subset (1,e)$. This allows us to bound the $n$ dependent piece in above integral as
$$\frac{1}{n+2} \le \int_0^1 (1-t)^{n+1} \frac{e^t - 1}{t} dt \le \frac{e}{n+2}$$
We also know for large $n$, $$H_n = \log n + \gamma + O\left(\frac1n\right)$$ Combine these, we can conclude
$$\sum_{n=0}^n (x_m - 1) = \log n + k + O\left(\frac1n\right) \quad\text{ where }\quad k = \gamma + \int_0^1 \frac{e^t-1}{t} dt$$ Using an CAS, we find the integral above has a closed form in terms of exponential integral ${\rm Ei}(x)$. The end result is
$$ k = \gamma + \left( {\rm Ei}(1) - \gamma\right) = {\rm Ei}(1) \approx 1.89511781635593675546652... $$