雜級數200811101134

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屬性

$\underset{n\to \infty }{\mathop{\lim }}\,\left( 1-\frac{1}{2^{2}} \right)\left( 1-\frac{1}{3^{2}} \right)\left( 1-\frac{1}{4^{2}} \right)\cdots \left( 1-\frac{1}{n^{2}} \right)=?$

$\underset{n\to \infty }{\mathop{\lim }}\,\left( 1-\frac{1}{2^{2}} \right)\left( 1-\frac{1}{3^{2}} \right)\left( 1-\frac{1}{4^{2}} \right)\cdots \left( 1-\frac{1}{n^{2}} \right)$
$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2^{2}-1}{2^{2}} \right)\left( \frac{3^{2}-1}{3^{2}} \right)\left( \frac{4^{2}-1}{4^{2}} \right)\cdots \left( \frac{n^{2}-1}{n^{2}} \right)$
$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{\left( 2+1 \right)\left( 2-1 \right)}{2\times 2} \right]\left[ \frac{\left( 3+1 \right)\left( 3-1 \right)}{3\times 3} \right]\left[ \frac{\left( 4+1 \right)\left( 4-1 \right)}{4\times 4} \right]\cdots \left[ \frac{\left( n+1 \right)\left( n-1 \right)}{n\times n} \right]$
$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n+1}{2n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1+\frac{1}{n}}{2}=\frac{1}{2}\#$