# A multiple integral question

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

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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

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.
Not at all$\phantom{}$ –  userNaN Feb 22 '13 at 19:11
Assuming what? $\phantom{}$ –  userNaN Feb 22 '13 at 20:30