Ta dễ dàng chứng minh được: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)
Thật vậy:
\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1>2n^2+2n=2n\left(n+1\right)\)Trở lại bài toán
\(A=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(A=\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+....+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(A< \dfrac{1}{2.1.\left(1+1\right)}+\dfrac{1}{2.2.\left(2+1\right)}+\dfrac{1}{2.3.\left(3+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\)
\(A< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{2n\left(n+1\right)}\)
\(A< \dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)\)
\(A< \dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(A< \dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)\)
\(A< \dfrac{1}{2}-\dfrac{1}{2n+2}< \dfrac{1}{2}\left(đpcm\right)\)
