$\frac{1}{9}+\frac{1}{25}+....+\frac{1}{(2n+1)^{2}}<\frac{1}{4} (n\geq 3)$
$\frac{1}{9}+\frac{1}{25}+....+\frac{1}{(2n+1)^{2}}<\frac{1}{4} (n\geq 3)$
Started By Quoc Tuan Qbdh, 15-06-2015 - 13:58
#1
Posted 15-06-2015 - 13:58
#2
Posted 15-06-2015 - 14:21
$\frac{1}{9}+\frac{1}{25}+....+\frac{1}{(2n+1)^{2}}<\frac{1}{4} (n\geq 3)$
từ đánh giá sau dễ dàng có được đpcm $\frac{1}{\left ( 2n+1 \right )^{2}}< \frac{1}{\left ( 2n+1 \right )^{2}-1}=\frac{1}{4n\left ( n+1 \right )}=\frac{1}{4}\left ( \frac{1}{n} -\frac{1}{n+1}\right )$
#3
Posted 15-06-2015 - 14:24
Edited by KySuBachKhoa, 15-06-2015 - 14:24.
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