1.
[TEX]\ln \left( {\frac{2}{1}.\frac{4}{3}..\frac{{2n}}{{2n - 1}}} \right) = \left( {\ln 2 - \ln 1} \right) + \left( {\ln 4 - \ln 3} \right) + .. + \left( {\ln \left( {2n} \right) - \ln \left( {2n - 1} \right)} \right)[/TEX]
Có
[TEX]\ln \left( {n + 1} \right) - \ln n > \frac{1}{{n + 1}}[/TEX]
[TEX] \to \ln \left( {\frac{2}{1}.\frac{4}{3}..\frac{{2n}}{{2n - 1}}} \right) > \frac{1}{2} + \frac{1}{4} + .. + \frac{1}{{2n}} = \frac{1}{2}\left( {1 + \frac{1}{2} + .. + \frac{1}{n}} \right)[/TEX]
Lại có
[TEX]\frac{1}{n} > \ln \left( {n + 1} \right) - \ln n[/TEX]
suy ra
[TEX] \to \ln \left( {\frac{2}{1}.\frac{4}{3}..\frac{{2n}}{{2n - 1}}} \right) > \frac{1}{2} + \frac{1}{4} + .. + \frac{1}{{2n}} = \frac{1}{2}\left( {1 + \frac{1}{2} + .. + \frac{1}{n}} \right) > \frac{1}{2}.\left( {\ln \left( {n + 1} \right) - 1} \right)[/TEX]
[TEX]\begin{array}{l}\to \lim \ln \left( {\frac{2}{1}.\frac{4}{3}..\frac{{2n}}{{2n - 1}}} \right) = + \infty \\\to \lim \left( {\frac{2}{1}.\frac{4}{3}..\frac{{2n}}{{2n - 1}}} \right) = + \infty \\ \to \lim \left( {\frac{1}{2}.\frac{3}{4}..\frac{{2n - 1}}{{2n}}} \right) = 0\end{array}[/TEX]
2/
[TEX]\lim VT = \lim \left( {\frac{1}{n}\sum\limits_{k = 0}^{n - 1} {\frac{1}{{\sqrt {1 - {{\left( {\frac{k}{n}} \right)}^2}} }}} - \frac{1}{n}} \right) = \int\limits_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} - \lim \frac{1}{n} = \arcsin x\left| {_0^1} \right. = \frac{\pi }{2}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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