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Bài 1:
Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
xong bn áp dụng lên trên lm tiếp
Bài 3:
theo bđt cô si ta có:
\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
Tương tự ta có :
\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
Cộng vế vs vế (1)(2)(3) ta có:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
\(\hept{\begin{cases}\frac{2}{2\sqrt{n}}< \frac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\frac{2}{2\sqrt{n}}>\frac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{cases}}\)
Từ đây ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2\left(\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+...+\sqrt{n}-\sqrt{n-1}\right)\)
\(=2\left(\sqrt{n}-0\right)=2\sqrt{n}\)
Tương tự ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\right)\)
\(=2\left(\sqrt{n+1}-1\right)>\sqrt{n}\)
Gọi \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}=A\)là A
Có \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{n}}\)
=> \(A>n.\frac{1}{\sqrt{n}}=\sqrt{n}\)(1)
Ta có: \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=2\left(\sqrt{n}+\sqrt{n-1}\right)\)
=> \(\frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
Khi đó: \(\frac{1}{\sqrt{1}}< 2\left(\sqrt{1}-\sqrt{0}\right)\)
\(\frac{1}{\sqrt{2}}< 2\left(\sqrt{2}-\sqrt{1}\right)\)
...
\(\frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
=> \(A< 2\left(\sqrt{n}-\sqrt{n-1}+...+\sqrt{1}\right)\)
=> \(A< 2\sqrt{n}\)(2)
Từ (1) và (2) => \(\sqrt{n}< A< 2\sqrt{n}\)
bạn Phạm Hữu Tiến, bạn mất dạy vừa thôi nha mình chưa làm j bạn, mình chỉ hỏi bài các bạn thôi, bạn không trả lời đc thì thôi chứ sao bạn lại nói tục như vậy?????????
\(VT< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n-1}=2-\frac{1}{n}\)
\(\frac{1}{1^2}=1,\frac{1}{2^2}< \frac{1}{1.2},\frac{1}{3^2}< \frac{1}{2.3},.....,\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=2-\frac{1}{n}\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}< 2-\frac{1}{n^2}\)
p/s: bài này giống vs toán lớp 6
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{n}{3^n}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{n}{3^{n-1}}\)
\(\Rightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{n}{3^{n-1}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{n}{3^n}\right)\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}-\frac{n}{3^n}\)
Đặt \(S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}\)
\(\Rightarrow3S=3+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}\)
\(\Rightarrow3S-S=\left(3+1+...+\frac{1}{3^{n-2}}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^{n-1}}\right)\)
\(\Rightarrow2S=3-\frac{1}{3^{n-1}}< 3\)
\(\Rightarrow2S< 3\)
\(\Rightarrow S< \frac{3}{2}\)
\(\Rightarrow2A< \frac{3}{2}\)
\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)
Thanks bn :))