So sánh: $C$ = $\frac{1}{1!}$ + $\frac{1}{2!}$ + $\frac{1}{3!}$ + … + $\frac{1}{2019!}$ với $\frac{7}{4}$ (Kí hiệu $n!$ = $1$.$2$.$3$.$4$……$n$)

0 thoughts on “So sánh: $C$ = $\frac{1}{1!}$ + $\frac{1}{2!}$ + $\frac{1}{3!}$ + … + $\frac{1}{2019!}$ với $\frac{7}{4}$ (Kí hiệu $n!$ = $1$.$2$.$3$.$4$……$n$)”

1. Giải thích các bước giải:

   Ta có:

   $C=\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+…+\dfrac{1}{2019!}$

   $\to C=1+\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+…+\dfrac{1}{1.2.3..2019}$

   $\to C<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{2018.2019}$

   $\to C<1+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+…+\dfrac{2019-2018}{2018.2019}$

   $\to C<1+1-\dfrac12+\dfrac12-\dfrac12+…+\dfrac1{2018}-\dfrac1{2019}$

   $\to C<2-\dfrac1{2019}$

   $\to C<2-\dfrac14$

   $\to C<\dfrac74$

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