Ta có \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)
=>\(\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}\left(\frac{1}{n\left(n+1\right)}\right)=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Áp dụng ta có \(\frac{1}{5}=\frac{1}{1^2+2^2}< \frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)\)
\(\frac{1}{13}=\frac{1}{2^2+3^2}< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)
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\(\frac{1}{2019^2+2020^2}< \frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)
=> \(VT< \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\right)=\frac{1}{2}\left(1-\frac{1}{2020}\right)< \frac{1}{2}\)(ĐPCM)
Câu hỏi của bạn sao ko thấy quy luật dãy nhỉ ?
mình đánh nhầm nhé sorry
1/3 thành 1/13
