Xét số hạng tổng quát:
\frac{1}{1+3+5+7+...+(2n-1)}=\frac{1}{\frac{(2n-1+1)n}{2}}=\frac{2}{2n.n}=\frac{1}{n^2}<\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})
Do đó:
A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}
\implies...