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B

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Đáp án B

$$\begin{gathered} S_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\mathrm{...}\left(1-\frac{1}{n^2}\right) \\ =\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\mathrm{...}\left(1-\frac{1}{n}\right)\right]\times \\ \left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{n}\right)\right] \end{gathered}$$

$\begin{array}{l}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\mathrm{...}\left(1-\frac{1}{n}\right)=\frac{1}{2}.\frac{2}{3}\mathrm{...}\frac{n-1}{n}=\frac{1}{n}\\ \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\mathrm{...}\left(1+\frac{1}{n}\right)=\frac{3}{2}.\frac{4}{3}\mathrm{...}\frac{n+1}{n}=\frac{n+1}{2}\\ \Rightarrow S_n=\frac{1}{n}.\frac{n+1}{2}=\frac{n+1}{2n}\\ \lim S_n=\lim \frac{n+1}{2n}=\lim \frac{1+\frac{1}{n}}{2}=\frac{1}{2}\end{array}$