đề thi toán cấp quốc gia

V

vy000

Ta có:

$\dfrac 9{n!} < \dfrac {(n-1)}{n!}=\dfrac 1{(n-1)!}-\dfrac 1{n!}$ Với $n >10,n \in Z$

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

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

$<\dfrac 1{9!}-\dfrac 1{10!}+\dfrac 1{10!}- \dfrac 1{11!}+\dfrac 1{11!}-\dfrac 1{12!}+...+\dfrac 1{999!}-\dfrac 1{1000!}$

$=\dfrac 1{9!}-\dfrac 1{1000!}$

$< \dfrac 1{9!}$
 
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