Cmr: \(2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}\)
Từ đó suy ra: \(2004<1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{1006009}}<2005\)
Những câu hỏi liên quan
1) CMR frac{1}{sqrt{1.1999}}+frac{1}{sqrt{2.1998}}+frac{1}{sqrt{3.1997}}+...+frac{1}{sqrt{1999.1}}ge1,9992) CMR frac{1}{1sqrt{2}+2sqrt{1}}+frac{1}{3sqrt{2}+2sqrt{3}}+...+frac{1}{95sqrt{94}+94sqrt{95}} 13) CMR frac{1}{2sqrt{1}}+frac{1}{3sqrt{2}}+frac{1}{4sqrt{3}}+...+frac{1}{left(n+1right)sqrt{n}} 24) CMR sqrt{n} frac{1}{sqrt{1}}+frac{1}{sqrt{2}}+frac{1}{sqrt{3}}+...+frac{1}{sqrt{n}} 2sqrt{n}
Đọc tiếp
1) CMR \(\frac{1}{\sqrt{1.1999}}+\frac{1}{\sqrt{2.1998}}+\frac{1}{\sqrt{3.1997}}+...+\frac{1}{\sqrt{1999.1}}\ge1,999\)
2) CMR \(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{95\sqrt{94}+94\sqrt{95}}< 1\)
3) CMR \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
4) CMR \(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)
1, CMR: \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\ge\frac{n}{n+1}\)
2, CMR: \(2\left(\sqrt{n-1}-1\right)< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}\)
3, CMR: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Bài 1: CMR
Bài 2: CMR
CMR \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)với n thuộc N*
Áp dụng cho S=\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
CMR 18<S<19
CMR\(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)với n thuộc N
Áp dụng CMR \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2500}}< 100\)
Ta có :
\(\hept{\begin{cases}\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\\sqrt{n+1}-\sqrt{n}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\end{cases}}\forall n\in N\)
Suy ra : \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\)
Đặt \(M=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2499}}+\frac{1}{\sqrt{2500}}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{2\sqrt{2500}}+\frac{1}{2\sqrt{2499}}+...+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{2}}+\frac{1}{2}\)
Áp dụng BĐT , ta có :
\(\frac{1}{2}M< \sqrt{2500}-\sqrt{2499}+\sqrt{2499}-\sqrt{2498}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}M< \sqrt{2500}-\sqrt{1}+\frac{1}{2}=50-\frac{1}{2}< 50\)
\(\Rightarrow M< 100\)
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CMR: \(\dfrac{1}{1\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{3\sqrt{4}}+...+\dfrac{1}{n\sqrt{n+1}}>2\) với n ϵ N*
CMR
\(\frac{43}{44}< \frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)
cmr với mọi n thuộc N* \(1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}< 2\sqrt{2}\)
CMR:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{3}}+....+\dfrac{1}{\left(n+1\right)\left(\sqrt{n}+n\sqrt{n+1}\right)}< 1\)
Cho n là số nguyên dương (n> hoặc = 2). CMR
\(\sqrt{n}< \dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}\)
Bạn ghi sai đề à? Số đầu tiên phải là \(\dfrac{1}{\sqrt{1}}\) chứ sao là \(\dfrac{1}{\sqrt{n}}\), mặc dù đề như vậy làm vẫn được nhưng chắc chẳng ai cho dãy quy luật kiểu đó
\(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}=2\left(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+...+\dfrac{1}{2\sqrt{n}}\right)\)
\(\Rightarrow A>2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)
\(\Rightarrow A>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\right)=2\left(\sqrt{n+1}-1\right)\)
Ta chứng minh \(2\left(\sqrt{n+1}-1\right)>\sqrt{n}\Leftrightarrow2\sqrt{n+1}>\sqrt{n}+2\)
\(\Leftrightarrow4\left(n+1\right)>n+4+4\sqrt{n}\Leftrightarrow3n>4\sqrt{n}\Leftrightarrow\sqrt{n}>\dfrac{4}{3}\)
\(\Leftrightarrow n>\dfrac{16}{9}\) (đúng với mọi \(n\ge2\) )
Vậy \(A>\sqrt{n}\)
- Ta chứng minh tiếp \(A< 2\sqrt{n}\)
\(A=1+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}=1+\dfrac{2}{2\sqrt{2}}+...+\dfrac{2}{2\sqrt{n}}\)
\(\Rightarrow A< 1+2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\right)\)
\(\Rightarrow A< 1+2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n-1}\right)\)
\(\Rightarrow A< 1+2\left(\sqrt{n}-1\right)=2\sqrt{n}-1< 2\sqrt{n}\) (đpcm)
Vậy: \(\sqrt{n}< \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}\)
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Nguyễn Việt Lâmtran nguyen bao quanBạch Tuyên NghiNguyễn Thanh Hằng help me
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