Limit of exponential integral $ \lim{t \rightarrow \infty}{(te^t\int_{t}^{\infty}\frac{e^{-s}}{s}ds)} $ 
Evaluate the following limit:
  \begin{equation}
\lim\limits_{t \rightarrow \infty}{(te^t\int\limits_{t}^{\infty}\frac{e^{-s}}{s}ds) := I}
\end{equation}


That's the Exponential integral, which is not an elementary function. Considering $ \frac{1}{s} < \frac{1}{t} $ we get $ I \le 1 $, and my intuition says it's actually the answer ($ \lim{I} = 1 $), but I can't prove that $ I \ge 1 $. Am I right?
 A: $$
\begin{align}
\lim_{t\to\infty}te^t\int_t^\infty\frac{e^{-s}}{s}\,\mathrm{d}s
&=\lim_{t\to\infty}t\int_0^\infty\frac{e^{-s}}{t+s}\,\mathrm{d}s&&s\mapsto t+s\\
&=\lim_{t\to\infty}\int_0^\infty\frac{e^{-s}}{1+s/t}\,\mathrm{d}s\\
&=\int_0^\infty e^{-s}\,\mathrm{d}s&&\begin{array}{}\text{Dominated}\\ \text{Convergence}\end{array}\\[8pt]
&=1
\end{align}
$$

If an asymptotic expansion is desired, we can use the fact that $\int_0^\infty s^n\,e^{-s}\,\mathrm{d}s=n!$ to get
$$
\begin{align}
te^t\int_t^\infty\frac{e^{-s}}{s}\,\mathrm{d}s
&=\int_0^\infty\frac{e^{-s}}{1+s/t}\,\mathrm{d}s\\
&=\int_0^\infty\left(1-\frac{s}{t}+\frac{s^2}{t^2}-\frac{s^3}{t^3}+\dots\right)\,e^{-s}\,\mathrm{d}s\\
&=0!-\frac{1!}{t}+\frac{2!}{t^2}-\frac{3!}{t^3}+\dots
\end{align}
$$
As with most asymptotic expansions, this converges nowhere.
A: Integrate by parts:
$$\begin{align}t e^{t} \int_t^{\infty} ds \frac{e^{-s}}{s} &= t e^t \left [ -\frac{e^{-s}}{s}\right]_{t}^{\infty} - t e^t \int_t^{\infty} ds \frac{e^{-s}}{s^2}\\ &= 1-t e^t \left [ -\frac{e^{-s}}{s^2}\right]_{t}^{\infty} + 2 t e^t \int_t^{\infty} ds \frac{e^{-s}}{s^3}\\ &= 1 -\frac{1}{t} + 2 t e^t \left [ -\frac{e^{-s}}{s^3}\right]_{t}^{\infty} - 6 t e^t \int_t^{\infty} ds \frac{e^{-s}}{s^4} \end{align}$$
etc.  Note that repeating integration by parts will produce terms in increasing powers of $1/t$.  You may then show the limit as being $1$.
