helping me

Ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
A=\frac{3}{2^{2}} +\frac{8}{3^{2}} +\frac{15}{4^{2}} +...+\frac{2023^{2} -1}{2023^{2}}\\
=\left( 1-\frac{1}{2^{2}}\right) +\left( 1-\frac{1}{3^{2}}\right) +\left( 1-\frac{1}{4^{2}}\right) +...+\left( 1-\frac{1}{2023^{2}}\right)\\
=2022-\left(\frac{1}{2^{2}} +\frac{1}{3^{2}} +\frac{1}{4^{2}} +...+\frac{1}{2023^{2}}\right)\\
\Longrightarrow A< 2022\ ( 1)
\end{array}$
Với mọi n ta có:
$\displaystyle \frac{1}{n^{2}} < \frac{1}{n( n-1)} =\frac{1}{n-1} -\frac{1}{n}$
Áp dụng công thức trên với n=2;3;4;...;2023 ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{1}{2^{2}} < \frac{1}{2-1} -\frac{1}{2} =1-\frac{1}{2}\\
\frac{1}{3^{2}} < \frac{1}{3-1} -\frac{1}{3} =\frac{1}{2} -\frac{1}{3}\\
\frac{1}{4^{2}} < \frac{1}{4-1} +\frac{1}{4} =\frac{1}{3} +\frac{1}{4}\\
....\\
\frac{1}{2023^{2}} < \frac{1}{2023-1} -\frac{1}{2023} =\frac{1}{2022} -\frac{1}{2023}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Longrightarrow \frac{1}{2^{2}} +\frac{1}{3^{2}} +\frac{1}{4^{2}} +...+\frac{1}{2023^{2}} < 1-\frac{1}{2} +\frac{1}{2} -\frac{1}{3} +...+\frac{1}{2022} -\frac{1}{2023}\\
\Longrightarrow \frac{1}{2^{2}} +\frac{1}{3^{2}} +\frac{1}{4^{2}} +...+\frac{1}{2023^{2}} < 1-\frac{1}{2023} < 1\\
\Longrightarrow 2022-\left(\frac{1}{2^{2}} +\frac{1}{3^{2}} +\frac{1}{4^{2}} +...+\frac{1}{2023^{2}}\right)  >2021\\
\Longrightarrow A >2021\ ( 2)
\end{array}$
Từ (1) và (2) có:  $\displaystyle 2021< A< 2022$
$\displaystyle \Longrightarrow $A không phải là số tự nhiên