So sánh $\frac{1}{2.9}$ + $\frac{1}{3.12}$ + $\frac{1}{4.15}$ + $\frac{1}{5.18}$ + … + $\frac{1}{2020.6063}$ với $\frac{1}{6}$
So sánh $\frac{1}{2.9}$ + $\frac{1}{3.12}$ + $\frac{1}{4.15}$ + $\frac{1}{5.18}$ + … + $\frac{1}{2020.6063}$ với $\frac{1}{6}$
\begin{array}{l}\quad\dfrac1{2.9}+\dfrac1{3.12}+\dfrac1{4.15}+\ldots+\dfrac1{2020.6063}\\=\dfrac3{3.2.9}+\dfrac3{3.3.12}+\dfrac3{3.4.15}+\ldots+\dfrac3{3.2020.6063}\\=\dfrac3{6.9}+\dfrac3{9.12}+\dfrac3{12.15}+\ldots+\dfrac3{6060.6063}\\=\dfrac16-\dfrac19+\dfrac19-\dfrac1{12}+\dfrac1{12}-\dfrac1{15}+\ldots+\dfrac1{6060}-\dfrac1{6063}\\=\dfrac16-\dfrac1{6063}<\dfrac16\\\text{⇒ $\dfrac1{2.9}+\dfrac1{3.12}+\dfrac1{4.15}+\ldots+\dfrac1{2020.6063}<\dfrac16$} \end{array}
$\begin{array}{l}\quad\dfrac1{2.9}+\dfrac1{3.12}+\dfrac1{4.15}+\ldots+\dfrac1{2020.6063}\\=\dfrac3{3.2.9}+\dfrac3{3.3.12}+\dfrac3{3.4.15}+\ldots+\dfrac3{3.2020.6063}\\=\dfrac3{6.9}+\dfrac3{9.12}+\dfrac3{12.15}+\ldots+\dfrac3{6060.6063}\\=\dfrac16-\dfrac19+\dfrac19-\dfrac1{12}+\dfrac1{12}-\dfrac1{15}+\ldots+\dfrac1{6060}-\dfrac1{6063}\\=\dfrac16-\dfrac1{6063}<\dfrac16\\\text{- Vậy $\dfrac1{2.9}+\dfrac1{3.12}+\dfrac1{4.15}+\ldots+\dfrac1{2020.6063}<\dfrac16$} \end{array}$