Prove that the lebesgue integrals $\lim_{n\to \infty} \int_{[0,1]^n} \frac{x_1^2+ \dots +x_n^2}{x_1+ \dots +x_n} \ d \mathrm{m}(\mathbf{x})=2/3$

Asked Jan 13 '14 at 03:50

Active Jan 13 '14 at 03:50

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I found this problem while studying for an analysis exam and have been puzzled ever since. If anybody has any hints on this, I would greatly appreciate it.

Prove that the lebesgue integrals $\lim_{n\to \infty} \int_{[0,1]^n} \frac{x_1^2+ \dots +x_n^2}{x_1+ \dots +x_n} \ d \mathrm{m}(\mathbf{x})=2/3$

asked Jan 13 '14 at 03:50

george

http://math.stackexchange.com/questions/623433/asymptotic-behaviour-of-a-multiple-integral-on-the-unit-hypercube – Jan 13 '14 at 03:59
Going to look that over now. Thanks a million :) – george Jan 13 '14 at 04:01
I think my solution is easiest to understand, but it uses probability theory. – Jan 13 '14 at 04:02
@ByronSchmuland Your solution does seem very concise. I wish I understood the argument in more rigor, though - the probability theory, a new field for me, seems quite interesting. – george Jan 16 '14 at 01:11

Prove that the lebesgue integrals $\lim_{n\to \infty} \int_{[0,1]^n} \frac{x_1^2+ \dots +x_n^2}{x_1+ \dots +x_n} \ d \mathrm{m}(\mathbf{x})=2/3$

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