CMR: \(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}< 2018\)
Tính: \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}\left(n\in N,n>0\right)\)
Help Nguyễn TrươngNguyễn Việt LâmKhôi Bùi Akai HarumaDƯƠNG PHAN KHÁNH DƯƠNG
Ta có \(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+n^2+2n+1+n^2}{n^2\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(n\left(n+1\right)+1\right)^2}{n^2\left(n+1\right)^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(=2018-\dfrac{1}{2018}< 2018\) (đpcm)
b/ Mọi tổng có dạng \(\sum\dfrac{1}{n}\) hay \(\sum\dfrac{1}{\sqrt{n}}\) gì đó đều ko tính ra kết quả cụ thể được, chỉ chứng minh chúng nằm trong khoảng nào đó thì được
