Asymptotic behavior of zeroes of $\ln(t)-t+n$ This is related to this question. Consider the sequence of functions $(f_n)_{n\geq 2}$ defined on $[1,\infty)$ by
$$f_n(t)=\ln(t)-t+n.$$
Those functions are decreasing on $[1,\infty)$ with $f_n(1)=n-1$ and $\lim_{t\to \infty}f_n(t)=-\infty$. So by intermediate value theorem, there exists a sequence of zeroes located in $[1,\infty)$, that is, $t_n>1$ and $f_n (t_n)=0$ for all $n\geq2$.
I am interested in the asymptotic behavior of the sequence $(t_n)_n$, specially finding an equivalent of this sequence at infinity.
Here's what I tried: $\ln(t_n)-t_n+n=0$ and $t_n>1$ is equivalent to 
$$t_n=e^{t_n-n}>1$$ which implies that $t_n>n$, so already the sequence diverges to infinity. Now again
$$t_n=e^{t_n-n}>n$$
which implies that
$$t_n>n+\ln(n)$$
repeating this argument we can find that
$$t_n>n+\ln(n+\ln(n+\ln(n+\dots)))$$
I don't know if this path leads to something. I still feel I need an upper bound to find an equivalent.
 A: The solutions to the equation
$$
\ln t - t + n = 0
$$
are given by
$$
t = -W(-e^{-n}),
$$
where $W$ is any branch of the Lambert W function. For $n \geq 1$ exactly two of these branches are real, namely
$$
t_{-1} = -W_{-1}(-e^{-n}) \qquad \text{and} \qquad t_0 = -W_0(-e^{-n}).
$$
Further, $t_{-1} \geq 1$ and $t_0 \leq 1$, so the solution you're interested in is $t_{-1}$.
If we define
$$
\begin{align}
&L_1(x) = \ln(-x), \\
&L_2(x) = \ln(-\ln(-x)),
\end{align}
$$
then the $-1$ branch of the $W$ function has a convergent (for $x < 0$ small enough) series expansion in powers of $L_1$ and $L_2$ given by
$$
W_{-1}(x) = L_1(x) - L_2(x) + \sum_{\ell = 0}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^\ell \begin{bmatrix} \ell + m \\ \ell + 1 \end{bmatrix}}{m!} L_1(x)^{-\ell-m} L_2(x)^m,
$$
where $\begin{bmatrix}p \\ q\end{bmatrix}$ is a Stirling cycle number.
Since $L_1(-e^{-n}) = -n$ and $L_2(-e^{-n}) = \ln n$, we get
$$
t_{-1} = n + \ln n - \sum_{\ell = 0}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^m \begin{bmatrix} \ell + m \\ \ell + 1 \end{bmatrix}}{m!} n^{-\ell-m} (\ln n)^m.
$$
This series can serve as an asymptotic expansion as $n \to \infty$, and the first few largest terms are
$$
t_{-1} = n + \ln n + \frac{\ln n}{n} - \frac{(\ln n - 2)\ln n}{2n^2} + O\!\left(\left(\frac{\ln n}{n}\right)^3\right).
$$
When $n = 10$ these terms give an approximate value of $t_{-1} \approx 12.52936$, whereas the true value is $t_{-1} \doteq 12.52796$.
A great reference is On the Lambert W Function by Corless et. al. [PDF Link].
