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I want to show $$\displaystyle\lim_{n\to \infty} \int_n^\infty \left(1+\frac{x}{n} \right)^n e^{-x} \, dx=0.$$

Hint is given ;

HINT : $\displaystyle\lim_{n\to \infty} u_n=0,$ where $u_n=\displaystyle\max_{x\geqq n}\left(1+\frac{x}{n}\right)^n e^{-x}.$

So far, I tried to do simply by $\epsilon-n.$

Let $\epsilon >0.$

Since $\lim u_n=0,$ there is $N\in \mathbb N$ s.t. $n\geqq N \Rightarrow u_n<\epsilon.$

When $n\geqq N$, $\left| \int_n^\infty \left(1+\frac{x}{n}\right)^n e^{-x} \, dx \right| \leqq \int_n^\infty \left| \left(1+\frac{x}{n}\right)^n e^{-x} \right| \, dx =\int_n^\infty \left(1+\frac{x}{n}\right)^n e^{-x} \, dx \leqq \int_n^\infty u_n \, dx < \int_n^\infty \epsilon \, dx.$

This fails because the upper limit of integral is $\infty.$ I don't know how I should handle the upper limit $\infty.$

I'd like you give me any idea.

daㅤ
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    Your integral can be written as $\int_\mathbb{R} (1+x/n)^ne^{-x}\chi_{(n,\infty)}$. This is an integral where the limits does not depends on $n$, so you can apply the dominate convergence theorem and interchange limit and integral. – Marcos Apr 05 '22 at 13:38
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    @Marcos Oh, good idea. Thank you. Lebesgue integral theory is so useful. – daㅤ Apr 05 '22 at 13:52
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    @Marcos: what is the dominating function in this case? – Mittens Apr 05 '22 at 15:07
  • @OliverDiaz, Oh, you are right. I deleted my dumb comment. I guess I need a cupful of espresso to wake up my brain properly... :( – Sangchul Lee Apr 05 '22 at 16:16

3 Answers3

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For $x \geq 1$, we have

$$ \log(1+x) \leq x - \frac{x^2}{2(1+x)} \leq \frac{3}{4}x $$

(The first inequality actually holds for all of $x \geq 0$ and can be proved by differentiation.) Using this, we get

$$ \left(1 + \frac{x}{n}\right)^n e^{-x} \mathbf{1}_{[n,\infty)}(x) \leq e^{-x/4}. $$

So by the dominated convergence theorem,

$$ \lim_{n\to\infty} \int_{n}^{\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \, \mathrm{d}x = \int_{0}^{\infty} \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n e^{-x} \mathbf{1}_{[n,\infty)}(x) \, \mathrm{d}x = 0. $$

Sangchul Lee
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  • Why is the last integral $\int_0^\infty$ ? $\ \int_n^\infty (1+x/n)^n e^{-x} dx=\int_{\mathbb R} (1+x/n)^n e^{-x} \mathbf{1}{[n,\infty)}(x) dx=\int{-\infty}^\infty (1+x/n)^n e^{-x} \mathbf{1}{[n,\infty)}(x) dx$ so I think the last integral is not $\int_0^\infty$ but $\int{-\infty}^\infty.$ – daㅤ Jun 12 '22 at 13:58
  • @SABAR, Both works, since $[n,\infty)\subseteq[0,\infty)\subset(-\infty,\infty)$. – Sangchul Lee Jun 12 '22 at 16:56
  • O.K. Thank you for the reply. – daㅤ Jun 13 '22 at 09:28
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For $x\geqslant0$, the function $x\mapsto(1+x/n)^ne^{-x/2}$ attains the maximum at $x=n$ (easy to check; the logarithmic derivative is $1/(1+x/n)-1/2$...). Thus $(1+x/n)^ne^{-x/2}\leqslant2^ne^{-n/2}$ and $$0\leqslant\int_n^\infty(1+x/n)^ne^{-x}\,dx\leqslant2^ne^{-n/2}\int_n^\infty e^{-x/2}\,dx=2(2/e)^n\underset{n\to\infty}{\longrightarrow}0.$$

metamorphy
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  • (+1) it seems your solution extend to $\int^\infty_n (1+x/n)^n e^{-ax}$ for $a>\frac12$ which is nice. (The obvious cases $a>1$ are very trivial of course) – Mittens Apr 05 '22 at 17:55
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    @OliverDiaz: Actually it extends to $a>\log 2$ (which is the best possible). – metamorphy Apr 06 '22 at 02:54
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Since the map $t\mapsto\Big(1+\frac{1}{t}\Big)^t$ is increasing in $t>0$ (see here), and $f_m(x)=\Big(1+\frac{x}{m}\Big)^me^{-x}$ is decreasing in $x>0$ for all $m$ ($f'_m(x)=-x\big(1+\frac{x}{m}\big)^{m-1}e^{-x}<0$), we have that

$$\begin{align}\mathbb{1}_{(n,\infty)}(x)\Big(1+\frac{x}{n}\Big)^ne^{-x}&=\sum^\infty_{m=n}\mathbb{1}_{(m,m+1]}(x)\Big(1+\frac{x}{n}\Big)^ne^{-x}\\ &\leq \sum^\infty_{m=n}\mathbb{1}_{(m,m+1]}(x)\Big(1+\frac{x}{m}\Big)^me^{-x}\\ &\leq\sum^\infty_{m=n}(2/e)^m\mathbb{1}_{(m,m+1]}(x) \end{align}$$

Integration yields $$\int^\infty_nf_n(x)\,dx\leq\sum^\infty_{m=n}(2/e)^m=(2/e)^n\frac{1}{1-\tfrac2e}\xrightarrow{n\rightarrow\infty}0$$

Mittens
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