Cho dãy số $({u_n})$ với ${u_n} = \left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)$. Khi đó $\lim {u_n}$bằng?
Trả lời bởi giáo viên
$\begin{array}{l}{u_n} = \left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right) = \left( {\dfrac{{{2^2} - 1}}{{{2^2}}}} \right).\left( {\dfrac{{{3^2} - 1}}{{{3^2}}}} \right)...\left( {\dfrac{{{n^2} - 1}}{{{n^2}}}} \right) = \dfrac{{\left( {{2^2} - 1} \right)\left( {{3^2} - 1} \right)...\left( {{n^2} - 1} \right)}}{{{2^2}{{.3}^2}...{n^2}}}\\ = \dfrac{{\left( {1.3} \right).\left( {2.4} \right).\left( {3.5} \right).\left( {4.6} \right)\,...\,\,\left[ {\left( {n - 1} \right).\left( {n + 1} \right)} \right]}}{{{2^2}{{.3}^2}...{n^2}}} = \dfrac{{n + 1}}{{2n}}\\ \Rightarrow \lim {u_n} = \lim \dfrac{{n + 1}}{{2n}} = \lim \dfrac{{1 + \dfrac{1}{n}}}{2} = \dfrac{1}{2}.\end{array}$
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