\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.....+\(\frac{1}{2017^2}\)<1
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Những câu hỏi liên quan
PN
7 tháng 10 2017
\(CM:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}+\frac{1}{2018^2}< \frac{3}{4}\)
\(=\frac{1}{2^2+3^2+4^2+...+2017^2+2018^2}\)
\(=\frac{1}{4044}\)
\(\Rightarrow\frac{1}{4044}< \frac{3}{4}\)
P/s: Ko chắc đâu nhé
7 tháng 10 2017
\(\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}\)
\(=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)
\(=>\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{3}{4}\)
FG
0
TT
1
CB
23 tháng 6 2017
VT<1/(3^2-1)+1/(5^2-1)+...+1/(2017^2-1)=1/(2.4)+1/(4.6)+...+1/(2016.2018)
=1/2 . (1/2-1/4+1/4-1/6+...+1/2016-1/2018)=1/4-1/(2.2018)<1/4
BK
1
BT
14 tháng 4 2017
Ta có: 1/22 < 1/1.2; 1/32<1/2.3; ....; 1/20172<1/2016.2017
=> 1/22+1/32+....+1/20172<1/1.2+1/2.3+...+1/2016.2017 = 1/1-1/2+1/2-1/3+....+1/2016-1/2017 = 1/1-1/2017= 1-1/2017 < 1
°> đpcm
24 tháng 4 2017
Ta có:
\(A=\frac{1}{2}+\frac{1}{2^2}+........+\frac{1}{2^{2017}}\)
\(\Rightarrow2A=1+\frac{1}{2}+.........+\frac{1}{2^{2016}}\)
Khi đó:
\(2A-A=\left(1+\frac{1}{2}+.....+\frac{1}{2^{2016}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{2017}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{2017}}\)
\(\Rightarrow A=\frac{2^{2017}-1}{2^{2017}}\)
\(\Rightarrow A< 1\)
VẬy: A < 1
24 tháng 4 2017
Ta có: 1/2+1/2^2+...+1/2^2017<1/1.2+1/2.3+...+1/2016.2017
1/2<1/1.2
1/2^2<1/2.3
..........
1/2^2017<1/2016.2017
7 tháng 3 2018
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đây ta có
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\)
T
8 tháng 3 2018
Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(\Leftrightarrow\sqrt{n}\left(\frac{1}{n}-\frac{1}{n1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\). Mà:
\(\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đó, ta có:
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\) (ĐPCM)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}< 1\)
1-\(\frac{1}{2017}\)<1