Lời giải:

Ta có:
$\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}=\frac{10}{11}<1$

Ta có điều phải chứng minh

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{15}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{15}\)

\(=1-\dfrac{1}{15}=\dfrac{14}{15}\)

Mà \(\dfrac{14}{15}< 1\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{15}< 3\)

a, Ta có :

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)

a) Đặt :

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

Ta thấy :

\(\dfrac{1}{2!}=\dfrac{1}{1.2}\)

\(\dfrac{1}{3!}=\dfrac{1}{1.2.3}\)

\(\dfrac{1}{4!}=\dfrac{1}{1.2.3.4}< \dfrac{1}{3.4}\)

.....................................

\(\dfrac{1}{100!}=\dfrac{1}{1.2.3..........100}< \dfrac{1}{99.100}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{99.100}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< 1-\dfrac{1}{100}\)

\(A< \dfrac{99}{100}< 1\)

\(\Rightarrow A< 1\rightarrowđpcm\)

b) Đặt :

\(B=\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+.............+\dfrac{9}{1000!}\)

Ta thấy :

\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)

\(\dfrac{9}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)

...................................................

\(\dfrac{9}{1000!}< \dfrac{1000-1}{1000!}=\dfrac{1}{999!}-\dfrac{1}{1000!}\)

\(\Rightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+............+\dfrac{1}{999!}-\dfrac{1}{1000!}\)

\(B< \dfrac{1}{9!}-\dfrac{1}{1000!}\)

\(\Rightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)

~ Chúc bn học tốt ~

N = \(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)

= \(\dfrac{1}{10.10}+\dfrac{1}{11.11}+\dfrac{1}{12.12}+...+\dfrac{1}{n.n}\)

=> N < \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+...+\dfrac{1}{\left(n-1\right).n}\)

=> N < \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(=>N< \dfrac{1}{9}-\dfrac{1}{n}\)

=> N < \(\dfrac{1}{9}\)

Vậy N < \(\dfrac{1}{9}\)

Ta có:

\(\dfrac{9}{n!}\)< \(\dfrac{n-1}{n!}\) = \(\dfrac{1}{(n-1)!} - \dfrac{1}{n!}\) với n > 10 (n thuộc Z)

\(\Rightarrow\) \(\dfrac{9}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!} \)

= \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!}\)

\(\Rightarrow\) \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{1}{10!} - \dfrac{1}{11!} + \dfrac{1}{11!} - \dfrac{1}{12!} + ....\)

= \(\dfrac{1}{9!} - \dfrac{1}{1000!}\)

\(\Rightarrow \) \(\dfrac{9}{10!} + \dfrac{9}{11!} + ...+ \dfrac{9}{1000!} < \dfrac{1}{9!}\)

Chúc bn hc tốt.

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