Convergence of the sequence $I_n=\int^\infty_n \big(1+\tfrac{x}{n}\big)^n2^{-x}\,dx$ This posting is motivated by the answer to the this posting.
Let
$$I(n;a)=\int^\infty_n \Big(1+\frac{x}{n}\Big)^n e^{-ax}\,dx,$$
where $0<a<\infty$ and $n\in\mathbb{N}$. It is clear that
$$I(n;a)\geq 2^n\int^\infty_ne^{-ax}\,dx=\frac{1}{a}\Big(\frac{2}{e^a}\Big)^n$$
From this, it follows that $\lim_{n\rightarrow\infty}I(n;a)=\infty$ for $0<a<\log 2$.
On the other hand, using the fact that $g_n(x)=\big(1+\tfrac{x}{n}\big)^ne^{-\tfrac{x}{2}}$ attains a (global) maximum at $x=n$ in the interval $(0,\infty)$
$$I(n;a)=\int^\infty_n g_n(x) e^{-(a-\frac12)x}\,dx\leq \frac{1}{a-\frac12}\Big(\frac{2}{e^a}\Big)^n$$
for all $a>\frac12$. From this, it follows that $\lim_{n\rightarrow\infty}I(n;a)=0$ for $a>\log(2)$ (of course $\log 2>\tfrac12$).
The inequalities presented above also show that for $a=\log 2$
$$\frac{1}{\log 2}\leq I_n=I(h;\log 2)\leq \frac{2}{2\log 2 -1}$$
The question(s) is (are): is $I_n$ convergent? if not what are the values of $\liminf_nI_n$ and $\limsup_nI_n$? I would be content with a prove or disprove of the former question; the later would be a treat.
 A: $\newcommand{\d}{\mathrm{d}}$
$\newcommand{\e}{\mathrm{e}}$
Slight shift in notation:
$$I_n(s) = \int_n^{\infty}\left(1+\frac{x}{n}\right)^n\e^{-(1/2 + s) x}\d x$$
Change variables $x = n + \tfrac{t}{s}$. Then
$$I_n(s)= \frac{\left(\e^{\ln 2 -(1/2 + s)}\right)^n}{s}\int_0^{\infty}(1+\tfrac{t}{2ns})^n\e^{-t/(2s) -t}\d t$$
Note that by Taylor's theorem $$n \ln \left(1+\frac{x}{n}\right) = x -\frac{x^2}{n}\int_0^1 \frac{u\,\d u}{1 + xu/n}$$
so that
$$-\frac{x^2}{2n}\leq n \ln \left(1+\frac{x}{n}\right) - x \leq 0\text{.}$$
Therefore
$$1 -\frac{x^2}{2n}\leq \left(1+\frac{x}{n}\right)^n\e^{-x}\leq 1\text{.}$$
Substituting,
$$ \frac{(\e^{\ln 2 -(1/2 + s)})^n}{s}
\int_0^{\infty}\left(1- \frac{t^2}{8s^2 n}\right)\e^{ -t}\d t \leq
 I_n(s) \leq 
\frac{(\e^{\ln 2-(1/2 + s)})^n}{s}
\int_0^{\infty}\e^{ -t}\d t\text{,}$$
viz.,
$$ \frac{1}{s}\left(1 -\frac{1}{4s^2n} \right)\leq (\e^{1/2 + s-\ln 2})^nI_n(s) \leq \frac{1}{s}$$
or
$$ 0 \leq \frac{1}{s} - (\e^{1/2 + s-\ln 2})^nI_n(s) \leq \frac{1}{4s^3 n} \text{.}$$
A: Let $X\sim \text{Exponential}(a)$. Then we have
\begin{align*}
I(n; a) = \frac{1}{a}\mathbb{E}\left[\left(1 + \frac{X}{n}\right)^n\bigg|X > n\right]\mathbb{P}(X>n)
\end{align*}
We can then use properties of the exponential distribution to derive
\begin{align*}
\mathbb{P}(X > n) &= e^{-an} \\
\mathbb{E}\left[\left(1 + \frac{X}{n}\right)^n\bigg|X > n\right] &=\mathbb{E}\left[\left(1 + \frac{X+n}{n}\right)^n\right] && \text{Memoryless Property of Exponential}\\
&= 2^n\mathbb{E}\left[\left(1 + \frac{\frac{1}{2}X}{n}\right)^n\right]
\end{align*}
Therefore at $a = \log(2)$,
\begin{align*}
I(n;\log(2)) = \frac{1}{\log(2)}\mathbb{E}\left[\left(1 + \frac{\frac{1}{2}X}{n}\right)^n\right]
\end{align*}
And so
\begin{align*}
\lim_{n\rightarrow \infty}I(n; \log(2)) &= \frac{1}{\log(2)}\mathbb{E}\left[\lim_{n\rightarrow \infty}\left(1 + \frac{\frac{1}{2}X}{n}\right)^n\right] && \text{Dominated convergence}\\
&= \frac{1}{\log(2)}\mathbb{E}\left[e^{\frac{1}{2}X}\right] \\
&= \frac{1}{\log(2)}\frac{\log(2)}{\log(2) - \frac{1}{2}} && \text{MGF of Exponential evaluated at 1/2} \\
&= \frac{1}{\log(2) - \frac{1}{2}}
\end{align*}
