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\(\frac{1}{2^2}< \frac{1}{1\cdot2}\\ \frac{1}{3^2}< \frac{1}{2\cdot3}\\ \frac{1}{4^2}< \frac{1}{3\cdot4}\\ ...\\ \frac{1}{n^2}< \frac{1}{\left(n-1\right)\cdot n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)\cdot n}\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\left(\text{với }n\in N;n\ge2\right)\)
1. D= 1/3 + 1/3.4 + 1/3.4.5 + 1/3.4.5....n < 1/2 + 1/3.4 + 1/4.5 + ...+ 1/ n.(n-1)
=> còn lại thì bạn có thể tự chứng minh
a/
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(A=2A-A=1-\frac{1}{2^{100}}< 1\)
b/
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)
a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=>đpcm
b) Từ công thức trên ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)
Đặt A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)< \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
=> A < 1 + (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/(n - 1) - 1/n)
=> A < 1 + (1 - 1/n)
=> A < 2 - 1/n
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{n-1}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{n-1}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}\right)\)
\(A=1-\frac{1}{2^n}< 1\)với mọi n -->Đpcm