Asymptotics of $\sum\limits_{n=1}^{\infty}a_n(t)$ when $t\to\infty$, where $a_1(t)=t$ and $a_{n+1}(t)=a_n(t)-1+e^{-a_n(t)}$ 
Let $\left\{a_n\right\}_{n\ge1}$ denote the sequence of functions from $\mathbb{R}^+$ to $\mathbb{R}$ defined, for every $t>0$, by $a_1(t)=t$ and, for every $n\ge1$, $$a_{n+1}(t)=a_n(t)-1+e^{-a_n(t)}.$$ Let $g(t)=\sum\limits_{n=1}^{\infty}a_n(t)$. Find
  $$\lim\limits_{t\to\infty}\ \left|g(t)-\tfrac{t(t+1)}{2}\right|.$$

What I tried: 
For every $t>0$, $\left\{a_n(t)\right\}_{n\ge1}$ is a decreasing sequence. And $a_{n+1}(t) = t-n+\sum\limits_{k=1}^{n}e^{-a_k(t)}<t$. 
From there I got no idea.
 A: This is not a solution. I am providing a plausibility argument to show the limit is of the form $\dfrac{e}{(e-1)^2} + \delta$, where $\delta$ is ''small'' (but unfortunately not too small).
Note if $a_k(t) > 0$, we have $a_k(t) - a_{k+1}(t) = 1 - \exp(-a_k(t)) > 0$ moreover since $e^x > 1 + x $ for $ x \neq 0$ we also have $a_{k+1} > 0.$ This implies $a_n(t)$ is a strictly decreasing positive sequence which converges to $a(t)$ where $a(t) = a(t) - 1 + e^{-a(t)}$ so $a(t) = 0.$
Also,$\lim_{n\to\infty}\dfrac{a_{n+1}(t)}{a^2_n(t)} = \lim_{n\to\infty}\dfrac{a_n(t) - 1 + \exp(-a_n(t))}{a_n^2(t)} = \dfrac{1}{2}.$ This suggests $a_n(t)$ converges to $0$ rapidly.
Lets analyze $(a_n(k))$ (denoted below as $(a_n)$ for simplicity) where $k$ is a large positive integer. So we have:
$$
\begin{align}
a_1 &= k, \\
a_2 &= k - 1 + e^{-k},\\
a_3 &= k - 2 + e^{-k} + e^{-((k - 1) + e^{-k})}\\
     &= k - 2 + e^{-k} + e^{-(k-1)} + e^{-(k-1)}({e^{-e^{-k}} - 1})\\
&\approx k - 2 + e^{-k} + e^{-(k-1)},\\
a_4 &\approx k - 3 + e^{-k} + e^{-(k-1)} + e^{-(k-2)},\\
\end{align}
$$
Assuming the approximation to continue, and assuming the terms $a_{k+2},\dots$ make small contribution to the sum, we have the following approximation to the sum: 
$$
\begin{align}
a_1 + a_2 + \dots + a_{k+1} &\approx  k + (k-1) + \dots + 0 + ke^{-k} + (k-1)e^{-(k-1)} + \dots e^{-1}\\
& \approx \dfrac{k(k+1)}{2} + \sum_{k=1}^{\infty} k e^{-k}\\
& \approx \dfrac{k(k+1)}{2} + \dfrac{e}{(e-1)^2} = \dfrac{k(k+1)}{2} +  .92067359420779231894.
\end{align}
$$ 
The above provides an approximation of the limit as $.92067359420779231894$. Using the multi-precision bc calculator I got an estimate of the true limit as $.941282470970473$, so the above approximation is within 2%.
