\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)
Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=>\(A>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
chứng tỏ rằng:
E=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)<\(\dfrac{3}{4}\)
Cho B=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{2014}{5^{2015}}\). Chứng tỏ rằng B<\(\dfrac{1}{16}\)
Cho:\(A=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{99^2}.Cm:\dfrac{1}{5}< A< \dfrac{1}{4}\)
1. Chứng tỏ rằng:
a) \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)
b) \(\dfrac{m}{a.\left(a+m\right)}=\dfrac{1}{a}-\dfrac{1}{a+m}\)
2. Tính
a) \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
b) \(\dfrac{5}{10.15}+\dfrac{5}{15.20}+...+\dfrac{5}{195.200}\)
c) \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{96.98}\)
chứng minh rằng
\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\) và B= 2
so sánh
A = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\)và \(B=\dfrac{1}{10}\)
mấy bạn ơi giúp mình câu này với
chứng minh rằng: \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{4010^2}< \dfrac{1}{2}\)
Chứng minh rằng :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}>\dfrac{25}{12}\)
Chúng minh:
S= \(\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
