khó qué đi mất , giải giúp mình zới

T

thaonguyenkmhd

$\frac{1}{2!} +\frac{2}{3!} +\frac{3}{4!} +... + \frac{99}{100!}

= (\frac{1}{1!} -\frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + (\frac{1}{3!} -\frac{1}{4!}) + ... + (\frac{1}{99!} -\frac{1}{100!})

=1 - \frac{1}{100!} <1$

Vậy $\frac{1}{2!} +\frac{2}{3!} +\frac{3}{4!} +... + \frac{99}{100!} < 1 $
 
H

harrypham

$$\begin{aligned} \frac1{2!}+\frac2{3!}+\frac3{4!}+ \cdots + \frac{99}{100!} & = \frac{2-1}{2!}+ \frac{3-1}{3!}+ \frac{4-1}{4!}+ \cdots + \frac{100-1}{100!} \\ & = \dfrac{1}{1!}- \frac{1}{2!}+ \dfrac{1}{2!}- \dfrac{1}{3!}+ \dfrac{1}{3!}- \dfrac{1}{4!}+ \cdots + \dfrac{1}{99!}- \frac{1}{100!} \\ & =1- \dfrac{1}{100!} <1 \end{aligned}$$
 
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