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$$\lim_{n \to \infty} \int_{0}^{n}\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}}dx$$

I thought about using the theorem of monotonic convergence and had $$f_n{(x)}=\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}} \lambda_{[0,n]}(x)$$ it is rising. but I don't know where it is valid to write: $$\lim_{n \to \infty} \int_{0}^{n}\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}}dx= \int_{0}^{\infty}\lim_{n \to \infty}\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}}dx=\int_{0}^{\infty}e^{\frac{x}{2}-3x}dx$$

Bozo Vulicevic
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2 Answers2

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The integral diverges. To see this, we can write

$$\int_0^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=\int_0^{n/3} \left(1-\frac{3x}n\right)^ne^{x/2}\,dx+\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx \tag 1$$

We will present two parts. In Part $1$, we will show that the first integral on the right-hand side of $(1)$ converges. In Part $2$, we will show that the second integral diverges.

PART $1$:

The first integral on the right-hand side of $(3)$ can easily be shown to converge. In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to establish that the exponential function satisfies the inequality

$$\left(1+\frac{x}{n}\right)^n\le e^x$$

for $x>-n$. Therefore, we assert that $\left(1-\frac{3x}n\right)^n\le e^{-3x}$ for $x<n/3$. We can then write

$$\begin{align} \int_0^{n/3}\left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\int_0^{\infty}\left(1-\frac{3x}n\right)^n\,\lambda_{[0,n]}(x)\,e^{x/2}\,dx\\\\ &\le \int_0^\infty e^{-5x/2}\,dx\\\\ &=\frac25 \end{align}$$

The dominated convergence theorem guarantees that

$$\begin{align} \lim_{n\to \infty}\int_0^{n/3}\left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\int_0^\infty \lim_{n\to \infty}\left(1-\frac{3x}n\right)^n\,\lambda_{[0,n]}(x)\,e^{x/2}\,dx\\\\ &=\int_0^\infty e^{-3x}\,(1)\,e^{x/2}\,dx\\\\ &=\frac25 \end{align}$$

PART $2$:

We will now analyze the second integral on the right-hand side of $(1)$. We enforce the substitution $x\to nx/3$ to reveal

$$\begin{align} \int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\frac{n}{3} \int_1^{3}\left(1-x\right)^ne^{nx/6}\,dx\\\\ &=(-1)^n\frac{n}{3}e^{n/6}\int_{0}^{2}x^ne^{nx/6}\,dx \end{align}$$

Since $\int_0^2 x^ne^{nx/6}\,dx\ge \frac{2^{n+1}-1}{n+1}$, then we have

$$\lim_{n\to \infty}\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx= \begin{cases} \infty&, n\,\,\text{even}\\\\ -\infty&, n\,\,\text{odd} \end{cases}$$

Therefore, the limit of interest does not exist and we cannot interchange the order of the limit and the integral.

Community
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Mark Viola
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  • I think the OP is correct. Baiscally, his idea is Lebesgue's Dominated Convergence Theorem but to use this theorem, he needs to find a function $g \in L^1$ such that the sequence of functions are $\leq g(x)$ for any $x$. Then the limit of the integral is $\frac{2}{5}$. – user398843 Feb 26 '18 at 21:19
  • @User398843 Have you read my answer? The issue is the part if the integrand forbwhich $x\ge n/3$. – Mark Viola Feb 26 '18 at 22:40
  • Sorry. I didn't. I just read the OP's question, then thought we can use LDCT to calculate it. I read your answer. It's pretty good. Is it because of the minus sign in $\left(1-\frac{3x}n\right)^n$, we need to discuss two cases? What about $\left(1+\frac{3x}n\right)^n$? – user398843 Feb 27 '18 at 02:22
  • Well, if the integrand were $\left(1+\frac{3x}n\right)^n e^{x/2}$, then the integral diverges also. – Mark Viola Feb 27 '18 at 03:06
  • Would you mind explaining a little bit why do we need to separate cases? I think it's easy to neglect discussing the cases – user398843 Feb 27 '18 at 03:56
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    We separate in cases because for $n/3< x\le n$, the absolute value of the integral exceeds $\frac13 n,e^{n/6}\frac{2^{n+1}-1}{n+1}$. – Mark Viola Feb 27 '18 at 04:24
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Let $f_n(x)=\left(1-\frac{3x}{n}\right)^ne^{x/2}$.

Then $$\lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)=\lim_{n\to\infty }f_n(x)\cdot \underbrace{\lim_{n\to\infty }\lambda_{[0,n]}(x)}_{=\lambda_{[0,\infty [}(x)}=\lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x).$$

Therefore $$\int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x)\mathrm d x=\int_0^\infty \lim_{n\to\infty }f_n(x)\mathrm d x.$$

Surb
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  • How do you justify interchanging the limit with the integral? – Mark Viola Feb 04 '16 at 17:29
  • I guess that if you ask the question, monotone convergence doesn't work as pretended the OP. I don't really have time to think about it right now, but let me know if we can't permute limit and integral. And if yes, feel free to modified my post to add more informations. @Dr.MV – Surb Feb 04 '16 at 17:44
  • This integral diverges in fact. So, we cannot interchange the order of the limit and the integral. – Mark Viola Feb 04 '16 at 18:01
  • @Dr.MV: Since the question was not about permuting limits and integral but really on how to get from $\int \lim_{n\to \infty }(f_n(x)\lambda_{[0,n]}(x))dx$ to $\int_0^\infty \lim_{n\to\infty }f_n(x)dx$, I will nevertheless let my post (I think it can be useful for the technical in the case where we can really permute limit and integral). But thank you for your comment, it's important to know that it's not possible here to permute limits and integral. – Surb Feb 04 '16 at 18:14
  • Surb, I thought the OP was explicitly asking whether the limit and the integral can be interchanged. If the upper limit were $n/3$ rather than $n$, then the answer would be "yes." However, as I showed in my answer, the integral (with upper limit $n$) diverges. So, everything you wrote is correct. But I thought I would try and present an answer to the question that I believe the OP is asking. ;-)) - Mark – Mark Viola Feb 04 '16 at 21:00