tìm lim

N

niemkieuloveahbu

[TEX]L=\lim_{n \to \infty}(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})[/TEX]

Ta có:

[TEX]\blue \Large L=\lim_{n \to \infty}(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})\\=\lim_{n \to \infty}\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{(n-1)(n+1)}{n^2}\\=\lim_{n \to \infty} \frac{1}{2}\frac{n+1}{n}=\frac{1}{2}[/TEX]
 
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