Hộ em câu này với aaaaaaaaaaa $\frac{1}{2^{2} }$ + $\frac{1}{4 ^{2} }$ + … + $\frac{1}{(2n)^{2} }$ < $\frac{1}{2}$ 11/11/2021 Bởi Mackenzie Hộ em câu này với aaaaaaaaaaa $\frac{1}{2^{2} }$ + $\frac{1}{4 ^{2} }$ + … + $\frac{1}{(2n)^{2} }$ < $\frac{1}{2}$
$\dfrac{1}{2^2}+\dfrac{1}{4^2}+….+\dfrac{1}{(2n)^2}$ $= \dfrac{1}{2^2}.\bigg(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+….+\dfrac{2}{n^2}\bigg)$ $ < \dfrac{1}{2^2}. \bigg( 1+ \dfrac{1}{1.2}+\dfrac{1}{2.3}+….+\dfrac{1}{(n-1).n}\bigg)$ $ = \dfrac{1}{2^2}.(2- \dfrac{1}{n} ) < \dfrac{1}{2}$ Bình luận
$\dfrac{1}{2^2}+\dfrac{1}{4^2}+….+\dfrac{1}{(2n)^2}$
$= \dfrac{1}{2^2}.\bigg(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+….+\dfrac{2}{n^2}\bigg)$
$ < \dfrac{1}{2^2}. \bigg( 1+ \dfrac{1}{1.2}+\dfrac{1}{2.3}+….+\dfrac{1}{(n-1).n}\bigg)$
$ = \dfrac{1}{2^2}.(2- \dfrac{1}{n} ) < \dfrac{1}{2}$