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22 tháng 6 2023

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

12 tháng 12 2019

Nếu \(n>0\Rightarrow\left(n-1\right)n\left(n+1\right)=n^3-n< n^3.\)

\(\Rightarrow VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2005.2006.2007}\)

\(\Rightarrow2.VT< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2005.2006.2007}\)

\(\Rightarrow2.VT< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2007-2005}{2005.2006.2007}\)

\(\Rightarrow2VT< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\)

\(\Rightarrow2.VT< \frac{1}{2}-\frac{1}{2006.2007}\Rightarrow VT< \frac{1}{4}-\frac{1}{2.2006.2007}< \frac{1}{4}\)

8 tháng 4 2022

`Answer:`

 \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)

a) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)

\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)

\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)

b) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)

\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)

\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)

\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)

\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)

27 tháng 7 2016

Ta có  \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

                                   \(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

áp dụng vào làm

14 tháng 1 2016

Ta có:3B\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+...+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}\)

B=\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+..+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}+\frac{1}{3}^{2005}\)

\(\Rightarrow\)2B=1-\(\frac{1}{3}^{2005}\)

\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)

\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}<\frac{1}{2}\)

\(\Rightarrow\)B<\(\frac{1}{2}\)