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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.

Mittens
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2 Answers2

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$\newcommand{\d}{\mathrm{d}}$ $\newcommand{\e}{\mathrm{e}}$ 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 $$ 0 \leq \frac{1}{s} - (\e^{1/2 + s-\ln 2})^nI_n(s) \leq \frac{1}{4s^3 n} \text{.}$$

Benjamin Wang
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K B Dave
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    Thanks for your posting @K.B.Dave. What change of variable you used to pass from the first equation to the second in the first identity of you posting? – Mittens Apr 06 '22 at 21:09
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    @JeanL. added the variable – K B Dave Apr 06 '22 at 21:20
  • @K.B.Dave: that was fantastic! not only did you obtained a different derivation of the convergence of $I(n;a=\tfrac12+s)$, $s>0$ but for $s=\log 2 -\tfrac12$ you get that $I_n$ (in my notation) converges to $\frac{1}{\log2-1/2}$. Thanks! – Mittens Apr 07 '22 at 00:03
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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*}

Tom Chen
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