Take the 2-minute tour ×
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.

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|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. –  Chris's sis 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

1 Answer 1

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|improve this answer
    
oh, nice!!! I just missed that! (+1) Thanks! :-) –  Chris's sis Feb 22 '13 at 19:10
    
Not at all$\phantom{}$ –  Norbert Feb 22 '13 at 19:11
1  
I love the simplicity of your answer! –  Chris's sis Feb 22 '13 at 19:12
    
Arent you assuming the consequent? –  CogitoErgoCogitoSum Feb 22 '13 at 20:05
3  
Assuming what? $\phantom{}$ –  Norbert Feb 22 '13 at 20:30

Your Answer

 
discard

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.