Mn giúp em với ạ : Cmr 1/2 + 1/3√2 + 1/4√3 +....+ 1/2005√2004 <2
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PM
1
KV
1
TM
5 tháng 7 2017
Xét dạng tổng quát:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\frac{1}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\)
\(< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)
\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)
\(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)
\(< 2\left(1-\frac{1}{\sqrt{2004}}\right)\)
\(< 2-\frac{2}{\sqrt{2004}}< 2\)
=>đpcm
LM
2
XO
4 tháng 9 2020
Ta có : B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)
=> 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
Khi đó 3B - B = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
=> 2B = \(1-\frac{1}{3^{2005}}\)
=> B = \(\frac{1}{2}-\frac{1}{3^{2005}.2}< \frac{1}{2}\left(\text{ĐPCM}\right)\)
NN
4 tháng 9 2020
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+........+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
Vì \(1-\frac{1}{3^{2005}}< 1\)\(\Rightarrow\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
hay \(B< \frac{1}{2}\)( đpcm )
HN
27 tháng 11 2017
Ta có:
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{matrix}\right.\)
Thế vô giải tiếp
VH
1
27 tháng 4 2017
Ta có :
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow\frac{2B}{2}=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\)
HT
1 tháng 6 2016
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{2005^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2004}-\frac{1}{2005}\)
\(=1-\frac{1}{2005}=\frac{2004}{2005}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2005^2}< \frac{2004}{2005}\left(\text{đ}pcm\right)\)
not hỉu