Ta có :
D = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
=> 2D = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
=> 2D - D = \(1-\frac{1}{2^{98}}\)
=> D = \(1-\frac{1}{2^{98}}\)\(<1\)
Chứng minh rằng :
\(y=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}
chứng minh rằng :
1/2!.3! + 2/1!.2!.3! + ... + 99/98!.99!.100! < 1
ChoN=1/2+(1/2)^2+(1/2)^3+(1/2)^4+......+(1/2)^98+(1/2)^99. Chứng minh B<1
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
Chứng minh rằng :100- ( 1+1/2+1/3+...+1/100)=1/2+2/3+3/4+...+99/100
chứng minh rằng :
1/ 1! . 2! + 2/ 1! . 2! . 3! + ... + 99/ 98! . 99! . 100! <1
Chứng minh 1/2+(1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^98+(1/2)^99 < 1
A=(1/2+1/2)^2+(1/2)^3+(1/2)^4+....+(1/2)^98+(1/2)^99
Chứng minh A<1