Пошаговое объяснение:
[tex]\lim_{n \to \infty} (1-\frac{1}{2^{2} } )(1-\frac{1}{3^{2} } )...(1-\frac{1}{n^{2} } )= \lim_{n \to \infty} \frac{(2^{2}-1)(3^{2}-1)...(n^{2}-1) }{(n!)^{2} }=\\ = \lim_{n \to \infty} \frac{1*3*2*4*3*5...*(n-1)(n+1)}{(n!)^{2}} =\lim_{n \to \infty} \frac{(n-1)!\frac{(n+1)!}{2} }{(n!)^{2}} =\\= \lim_{n \to \infty} \frac{n+1}{2n}= \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})=\frac{1}{2}[/tex]
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Пошаговое объяснение:
[tex]\lim_{n \to \infty} (1-\frac{1}{2^{2} } )(1-\frac{1}{3^{2} } )...(1-\frac{1}{n^{2} } )= \lim_{n \to \infty} \frac{(2^{2}-1)(3^{2}-1)...(n^{2}-1) }{(n!)^{2} }=\\ = \lim_{n \to \infty} \frac{1*3*2*4*3*5...*(n-1)(n+1)}{(n!)^{2}} =\lim_{n \to \infty} \frac{(n-1)!\frac{(n+1)!}{2} }{(n!)^{2}} =\\= \lim_{n \to \infty} \frac{n+1}{2n}= \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})=\frac{1}{2}[/tex]