Đáp án: $A$ < $\dfrac{2017}{1152}$
Giải thích các bước giải:
Ta có :
$A$ = $\dfrac{1}{1!}$ + $\dfrac{1}{2!}$ + $\dfrac{1}{3!}$ + ... + $\dfrac{1}{2018!}$
$A$ = $\dfrac{1}{1}$ + $\dfrac{1}{1.2}$ + $\dfrac{1}{1.2.3}$ + ... + $\dfrac{1}{1.2.3...2018}$
Mà
$\dfrac{1}{5!}$ = $\dfrac{1}{1.2.3.4.5}$ < $\dfrac{1}{3.4.5}$
`1/6!` = `1/1.2.3.4.5.6` < `1/4.5.6`
. .................................................................................................
$\dfrac{1}{2018!}$ = $\dfrac{1}{1.2.3....2018}$ < $\dfrac{1}{2016.2017.2018}$
`⇒` $A$ < `1` + `1/2` + `1/1.2.3` + `1/2.3.4` + `1/3.4.5` +`1/4.5.6` ..... +$\dfrac{1}{2016.2017.2018}$
Ta lại có : $\dfrac{1}{n(n+1)(n+2)}$ = `1/2` . ( $\dfrac{1}{n(n+1)}$ - $\dfrac{1}{(n+1)(n+2)}$ )
⇒ $A$ < `1` + `1/2` . ( `1/1.2` - `1/2.3` + `1/2.3` - `1/3.4` + .... + `1/2016.2017` - `1/2017.2018` )
⇒ $A$ < `1` + `1/2`. ( `1/1.2` - `1/2017.2018` ) < `1` + `1/2` + `1/1.2` . `1/1.2`
⇒ $A$ < `1` + `1/2`. ( `1/1.2` - `1/2017.2018` ) < `7/4`
Mà `7/4` = `2016/1152` < `2017/1152`
⇒ $A$ < `7/4` < `2017/1152`
⇒ $A$ < `2017/1152`
Vậy $A$ < `2017/1152`
