CMR: \(\frac{1}{2\sqrt[3]{1}}+\frac{1}{3\sqrt[3]{2}}+\frac{1}{4\sqrt[3]{3}}+...+\frac{1}{\left(n+1\right)\sqrt[3]{n}}\) với mọi \(n\varepsilonℕ^∗\)
Những câu hỏi liên quan
Bài 1: CMR
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+........+\frac{1}{\left(n+1\right)\sqrt{n}}>2,n\varepsilonℕ^∗\)
Bài 2: Cho S= \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{3\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
CMR S<\(\frac{1}{2}\)
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: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
14 tháng 5 2020 lúc 21:34
CMR : với mọi số tự nhiên n > 1, ta có :
a) \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)
b) \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
a) Ta có \(\frac{1}{n+k}>\frac{1}{2n}\)với k=1;2;...;n-1
=> \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}>\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}=\frac{n}{2n}=\frac{1}{2}\)
Mặt khác ta có \(\frac{1}{n+k}+\frac{1}{n\left(+\left(n+1-k\right)\right)}< \frac{3}{2n}\)
\(\Leftrightarrow3k^2+3nk+n+3k\forall k=1;2;...;n\)
Với k=1 ta có \(\frac{1}{n+1}+\frac{1}{n+n}< \frac{3}{2n}\)
Với k=2 ta có \(\frac{1}{n+2}+\frac{1}{n+\left(n-1\right)}< \frac{3}{2n}\)
..........................................
Với k=n ta có \(\frac{1}{n+n}+\frac{1}{n+1}< \frac{3}{2n}\)
Cộng từng vế của 2 BĐT trên ta được
\(2\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right)< \frac{3}{2n}+\frac{3}{2n}+....+\frac{3}{2n}=\frac{3n}{2n}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)(đpcm)
Không cần chứng minh \(\frac{1}{2}< \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\)
b) Ta có \(\frac{1}{\sqrt{k}}=\frac{1}{2\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)
Khi cho k=1,.....,n ta có:
\(1>2\left(\sqrt{2}-1\right)\)
\(\frac{1}{\sqrt{2}}>2\left(\sqrt{3}-\sqrt{2}\right)\)
................................
\(\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
Cộng từng vế BĐT trên ta có \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có đpcm
Xem thêm câu trả lời
Chứng minh rằng với mọi số nguyên dương n ta đều có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+\frac{1}{5\sqrt{4}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Xét số hạng tổng quát ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\sqrt{n}\cdot\frac{2}{\sqrt{n}}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài tập, ta có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+...+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
\(=2-\frac{2}{\sqrt{n+1}}< 2\left(đpcm\right)\)
a) CMR: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\) với \(n\in N\)*
b) tính \(B=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+......+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
a/ Quy đồng vế phải, hình như lộn mẫu cuối là căn 2 của (n+1) mới đúng
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\(VP=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}}{\sqrt{n+1}.\sqrt{n}}\)
\(=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right).\sqrt{n+1}.\sqrt{n}}=\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=VT\)
\(B=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=1-\frac{1}{\sqrt{25}}\)
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Cmr: \(\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...\frac{1}{\left(n+1\right)\sqrt{n}}< \sqrt{2}\)
CMR: Với mọi n \(\in Z\)+, ta có : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}<2\)
Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)=\(\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(1-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{n}-\frac{1}{\sqrt{n+1}}\right)\)
nên \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+.....+\frac{1}{\left(n+1\right)\sqrt{n}}\)\(< 2\left(\left(\frac{1}{n}-\frac{1}{\sqrt{n+1}}\right)+...+\left(3\sqrt{2}-2\right)+\left(2-1\right)\right)\) = 2
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CMR:
a, \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}>\frac{n}{n+1}\)
b, \(2\left(\sqrt{n-1}-1\right)< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n-1}\)
c, \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Mấy bài này đã có người làm rồi nhé bạn vào câu hỏi tương tự mà xem.
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