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a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2018}{2019!}\\ =\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{2018!}-\dfrac{1}{2019!}\right)\\ =1-\dfrac{1}{2019!}< 1\)
b) \(\dfrac{1\cdot2-1}{2!}+\dfrac{2\cdot3-1}{3!}+...+\dfrac{999\cdot1000-1}{1000!}\\ =\dfrac{1\cdot2}{2!}-\dfrac{1}{2!}+\dfrac{2\cdot3}{3!}-\dfrac{1}{3!}+...+\dfrac{999-1000}{1000!}-\dfrac{1}{1000!}\\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+...+\dfrac{1}{999!}+\dfrac{1}{1000!}\\ =1+1-\dfrac{1}{1000!}\\ =2-\dfrac{1}{1000!}< 2\)
Đặt A= 1/1.2 + 1/2.3 + 1/3.4+...+ 1/999.1000
=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000
=1-1/1000
=999/1000
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{999.1000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(=1-\dfrac{1}{1000}\)
\(=\dfrac{999}{1000}\)
Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
= \(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+....+\frac{99.100}{100!}-\frac{1}{100!}\)
= \(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
= \(\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}\right)\)
= \(1+1-\frac{1}{99!}\)
= \(2-\frac{1}{99!}
