Chứng minh \(S=1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+n}< 2\)
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KV
30 tháng 10 2023
S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²
⇒ S/3 = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³
⇒ 2S/3 = S - S/3
= (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²) - (1/3² +1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³)
= 1/3 - 1/3²⁰²³
⇒ S = (1/3 - 1/3²⁰²³) : 2/3
= (1 - 1/3²⁰²²) : 2
Lại có: 1 - 1/3²⁰²² < 1
⇒ S < 1/2
30 tháng 3 2017
1.
Ta có:
Vì b+1-b=1=>\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{1}{b.\left(b+1\right)}\)<\(\dfrac{1}{b.b}\)(1)
Vì b-(b-1)=1=>\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{1}{b.\left(b-1\right)}\)>\(\dfrac{1}{b.b}\)(2)
Từ (1) và (2)=>\(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b.b}< \dfrac{1}{b-1}-\dfrac{1}{b}\)
Câu 2 bạn hỏi bạn Bùi Ngọc Minh nhé PR cho nó
30 tháng 3 2017
Bài 2:
Ta có:S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{9^2}=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)
S>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\left(1\right)\)
S<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
8 tháng 4 2017
Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)
Ta lại có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)
Từ (1) và (2) suy ra đpcm.
23 tháng 6 2018
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
24 tháng 6 2018
Câu 1:
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\\ \Leftrightarrow2\sqrt{x-a}+2\sqrt{y-b}+2\sqrt{z-c}=x+y+z\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}=0\\ \Leftrightarrow x+y+z-2\sqrt{x-a}-2\sqrt{y-b}-2\sqrt{z-c}+3-a-b-c=0\\ \Leftrightarrow\left[\left(x-a\right)-2\sqrt{x-a}+1\right]+\left[\left(y-b\right)-2\sqrt{y-b}+1\right]+\left[\left(z-c\right)-2\sqrt{z-c}+1\right]=0\\ \Leftrightarrow\left(\sqrt{x-a}-1\right)^2+\left(\sqrt{y-b}-1\right)^2+\left(\sqrt{z-c}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}-1=0\\\sqrt{y-b}-1=0\\\sqrt{z-c}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-a}=1\\\sqrt{y-b}=1\\\sqrt{z-c}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-a=1\\y-b=1\\z-c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=a+1\\y=b+1\\z=c+1\end{matrix}\right.\)Vậy \(\left\{x;y;z\right\}=\left\{a+1;b+1;c+1\right\}\)
24 tháng 6 2018
Câu 2:
\(\text{ a) Ta có }:\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\\ =\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\left(1\right)\)
\(\text{Lại có: }\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}-\sqrt{n}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow2\left(\sqrt{n+1}-n\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
b) Áp dụng bất đảng thức ở câu a:
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\\ >2\left(\sqrt{101}-\sqrt{100}\right)+...+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{100}+...+\sqrt{4}-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\right)\\ =2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-1\right)=2\left(10-1\right)=18\left(3\right)\)
\(\Rightarrow S=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 2\left(\sqrt{100}-\sqrt{99}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-\sqrt{0}\right)\\ =2\left(\sqrt{100}-\sqrt{99}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}+\sqrt{1}\right)\\ =2\cdot\sqrt{100}=2\cdot10=20\left(4\right)\)
Từ \(\left(3\right)\) và \(\left(4\right)\Rightarrow18< S< 20\)
1 tháng 1 2019
a)
+) Ta có: \(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\) \(=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\) (1)
+) Ta có:
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}< \dfrac{2}{\sqrt{n}+\sqrt{n-1}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\) \(=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-\left(n-1\right)}\)
\(=2\left(\sqrt{n}-\sqrt{n-1}\right)\) (2)
Từ (1) và (2) ⇒ đpcm
Học toán vui vẻ! 
25 tháng 8 2017
Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
26 tháng 8 2017
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
16 tháng 5 2017
Ta có:
S = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}=1-\dfrac{1}{20}< 1\)
Vậy S<1
17 tháng 4 2018
Ta có:
S = 12+122+123+...+1220<11.2+12.3+13.4+...+119.20=1−120<112+122+123+...+1220<11.2+12.3+13.4+...+119.20=1−120<1
Vậy S<1
*chứng minh:
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+n}< 1\)
Xét thừa số tổng quát:
\(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)
\(\Rightarrow\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+n}\)
\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{n\left(n+1\right)}=2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=1-\dfrac{1}{2\left(n+1\right)}< 1\)
Điều phải chứng minh ~
p/s: xem lại,tối t hay ngáo lắm,nghĩ thế này gõ thế khác đó ^^
Èo