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Trước hết bạn chứng minh bằng quy nạp hệ thức sau
$1+3+\cdots+(2n-1) = \sum_{k=1}^n(2k-1)=n^2$
Ta có $\mathop {\lim }\limits_{n \to \infty }(\frac{1+3+5+...+2n-1}{n+1}$-$\frac{2n+1}{2})=\mathop {\lim }\limits_{n \to \infty }(\frac{n^2}{n+1}-\frac{2n+1}{2})=\mathop {\lim }\limits_{n \to \infty }(\frac{n^2}{n+1}-\frac{2n+1}{2})$
$=\mathop {\lim }\limits_{n \to \infty }-\frac{3n+1}{2n+2}=-\frac{3}{2}$
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