Chứng minh rằng:
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2008^2}<1\)
b) \(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}>\frac{13}{21}\)
Mong mọi người giúp em với ạ!
a) Cho A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}\)Chứng minh A<\(\frac{3}{4}\).
b) Chứng minh rằng:A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{99}}< \frac{1}{2}\)
b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3A-A=\(1-\frac{1}{3^{99}}\)
2A=\(1-\frac{1}{3^{99}}\)
vì 2A<1
=> A<\(\frac{1}{2}\)
Chứng minh rằng \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.............+\frac{1}{2009\sqrt{2008}}< 2\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{\sqrt{n^2}}-\frac{1}{\sqrt{\left(n+1\right)^2}}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(< \left(1+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán ta được
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2009\sqrt{2008}}\)
\(=2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2009}}\right)< 2\)
chứng minh rằng:a/ \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\) \(\frac{1}{2}\)
b/\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)\(\frac{1}{2}\)
nhanh thì tích
chậm thì thôi
Cho A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)
Chứng minh : \(\frac{2017}{2018} > A > \frac{2008}{2018} \)
Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
Xét B = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
=\(1-\frac{1}{2018}\)
Xét : \(\frac{2018}{2018}=1\)=) B < 1
khoan hình như sai đề
Chứng minh : \(\frac{1}{1}-\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{5}-\frac{1}{8}+\frac{1}{13}-\frac{1}{21}+\frac{1}{34}-\frac{1}{55}+...< \frac{3}{10}\).
Chứng minh rằng:
\(K=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}<\frac{3}{4}\)
Chứng minh rằng
\(K=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}<\frac{3}{4}\)
Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2002^2}\)
Chứng minh rằng A<\(\frac{1505}{2008}\)
Chứng minh :
A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2008^2}< 1\)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}\left(1\right)\)
Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=1-\frac{1}{2008}< 1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\) (đpcm)