\( \lim _ { n \rightarrow \infty } ( \frac { 1 + 3 + 5 + 7 + \cdots + ( 2 n - 1 ) } { n + 1 } - \frac { 2 n + 1 } { 2 } ) = - \frac { 3 } { 2 } \)

The sum of the series \(1 + 3 + 5 + 7 + \cdots + (2n - 1)\) represents the sum of the first \(n\) odd numbers, which is known to be \(n^2\). So, we can replace the numerator with \(n^2\), giving us: \[ \lim _ { n \rightarrow \infty } \left( \frac{n^2}{n + 1} - \frac{2n + 1}{2} \right). \] Now for the limit, we can simplify the expression: 1. The first term simplifies to \(\frac{n^2}{n + 1} = \frac{n^2}{n(1+\frac{1}{n})} \approx n(1 - \frac{1}{n}) = n - 1\) as \(n \to \infty\). 2. The second term is \(\frac{2n + 1}{2} = n + \frac{1}{2}\). Now, substituting these back into the limit gives us: \[ \lim_{n \to \infty} \left( n - 1 - \left(n + \frac{1}{2}\right) \right) = \lim_{n \to \infty} \left( -1 - \frac{1}{2} \right) = -\frac{3}{2}. \] Thus, indeed: \[ \lim _ { n \rightarrow \infty } \left( \frac { 1 + 3 + 5 + 7 + \cdots + ( 2 n - 1 ) } { n + 1 } - \frac { 2 n + 1 } { 2 } \right) = -\frac{3}{2}. \]