Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Proving that $$\lim_{n\to\infty}\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{n \text{ times}}\frac{1}{(x_1\cdot x_2\cdots x_n)^2+1} \mathrm{d}x_1\cdot\mathrm{d}x_2\cdots\mathrm{d}x_n=1$$

share|cite|improve this question
looks like we can make the use of Taylor's expansion this will turn out to be $ \lim_{n\to \infty} \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^n}$ which converges to $1$. Nice Question as always (+1) – Santosh Linkha Feb 23 '13 at 7:36
@experimentX: thank you! :-) Does Taylor's expansion work here? Let me check that. – user 1618033 Feb 23 '13 at 7:38
looks like it wasn't Taylor's expansion ... just expansion of $1/(1+x^2)$ haha .. pardon my math vocabulary. – Santosh Linkha Feb 23 '13 at 8:04
up vote 12 down vote accepted

Since $$ \int\limits_{0}^{1}\ldots\int\limits_{0}^{1}dx_n\ldots dx_1=1 $$ it is enough to show that $$ \lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}\left(1-\frac{1}{(x_1\cdot\ldots\cdot x_n)^2+1}\right)dx_n\ldots dx_1=0 $$ which is equivalent to $$ \lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}\frac{(x_1\cdot\ldots\cdot x_n)^2}{(x_1\cdot\ldots\cdot x_n)^2+1}dx_n\ldots dx_1=0 $$ Now note that $$ \begin{align} 0&\leq \lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}\frac{(x_1\cdot\ldots\cdot x_n)^2}{(x_1\cdot\ldots\cdot x_n)^2+1}dx_n\ldots dx_1\\ &\leq \lim\limits_{n\to\infty}\int\limits_{0}^{1}\ldots\int\limits_{0}^{1}(x_1\cdot\ldots\cdot x_n)^2dx_n\ldots dx_1\\ &=\lim\limits_{n\to\infty}\left(\frac{1}{3}\right)^n=0 \end{align} $$ And the result follows.

share|cite|improve this answer
oh, nice!!! I just missed that! (+1) Thanks! :-) – user 1618033 Feb 22 '13 at 19:10
Not at all$\phantom{}$ – Norbert Feb 22 '13 at 19:11
I love the simplicity of your answer! – user 1618033 Feb 22 '13 at 19:12
Arent you assuming the consequent? – CogitoErgoCogitoSum Feb 22 '13 at 20:05
Assuming what? $\phantom{}$ – Norbert Feb 22 '13 at 20:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.