# Compute $\lim\limits_{n\to\infty}\int_0^1\cdots\int_0^1 \max\{x_1,\ldots ,x_n\}\,dx_1\cdots dx_n.$ [closed]

Problem: Compute $$\lim_{n\to\infty}\int_0^1\cdots\int_0^1 \max\{x_1,\ldots ,x_n\} \, dx_1\cdots dx_n.$$

What I think: Let $x_1$ be the maximum of the set $\{ x_1,\ldots ,x_n \}$ so we have $1\over2$ and $n-1$ number $1.$ The $1^\infty$ reminds me of $\left(1+\frac 1 n \right)^n=e.$ On the contrary, if $x_n=n$ so how is it going the result.

## closed as off-topic by Andrés E. Caicedo, Xander Henderson, Leucippus, Did, JonMark PerrySep 5 '17 at 5:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Leucippus, Did, JonMark Perry
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• You've been around for two years and posted dozens of questions. You should know by now that posting a question like this is going to get a negative response.... – user296602 Sep 3 '17 at 20:38
• @HamedBaghalGhaffari You got a negative response for the same reason as this question from a week ago or this one from two weeks ago. Please read the closure reasons on those posts, and improve this question accordingly. It's got 3 votes to close already. – user296602 Sep 3 '17 at 20:41
• Since it's not homework, perhaps adding something to the question about how you came to think of this question might satisfy those who want more context. – Michael Hardy Sep 3 '17 at 20:56
• It's because this isn't a site that will solve your homework for you, and some effort or examples of your own thoughts must be shown. For more information, please read how to ask a good question. Best of luck to you. – let's have a breakdown Sep 3 '17 at 23:15
• @MichaelHardy "With newbie users they ought to explain that here in comments when they vote to close it for that reason" This is your own, very peculiar, interpretation, which you know very well has been refuted tons of times. Please stop mixing honest advice with propaganda. – Did Sep 4 '17 at 8:51

$$n\operatorname{-\text{dimensional volume}}\Big(\{ (x_1,\ldots,x_n) : \max\{x_1,\ldots,x_n\} <c \}\Big) = c^n,$$ and therefore $$\operatorname{volume}\Big( \{(x_1,\ldots,x_n) : \max \ge c\} \Big) = 1-c^n,$$ and so $$I_n = \text{integral} \ge c(1-c^n).$$ But clearly $I_n \le 1.$
So $$1 \ge I_n \ge c(1-c^n) \quad \text{ for all } c\in(0,1).$$
Hence $$1 \ge \lim_n I_n \ge \lim_n c(1-c^n) = c, \quad \text{for all } c\in(0,1).$$ $$\text{For all } c\in(0,1), \quad 1 \ge \lim_n I_n \ge c$$ So $\lim\limits_n I_n = 1.$
• By the way, the integral itself, without the limit as $n\to\infty,$ is $\dfrac n {n+1}. \qquad$ – Michael Hardy Sep 3 '17 at 21:07
I would look at this probabilistically: You have $n$ independent random variables uniformly distributed between $0$ and $1.$ If $n$ is large, then the probability that their maximum is near $1$ is very high. The integral is the expected value of the maximum. As $n$ grows, that expected value approaches $1.$