Determine the convergence or divergence of the sequence with the given \( n \)th term. If the sequence converges, find its limit. (If the quantity diverges, enter DIVERGES.) \[ a_{n}=(-1)^{n}\left(\frac{n}{n+2}\right) \]

To determine the convergence or divergence of the sequence \( a_n = (-1)^{n} \left( \frac{n}{n+2} \right) \), we first analyze the absolute value of the term without the alternating factor: \[ \left| a_n \right| = \frac{n}{n+2} \] As \( n \) approaches infinity, we examine the limit of this absolute value: \[ \lim_{n \to \infty} \frac{n}{n+2} = \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n}} = 1 \] Since this limit is equal to 1 and not approaching 0, we know that \( \left| a_n \right| \) does not converge to 0. Now, let's consider the behavior of the sequence \( a_n \): - When \( n \) is even, \( a_n \) takes the positive value \( \frac{n}{n+2} \). - When \( n \) is odd, \( a_n \) takes the negative value \( -\frac{n}{n+2} \). Thus, as \( n \) increases, the sequence oscillates between values that approach 1 and -1: \[ \lim_{n \to \infty} a_{2n} = 1 \quad \text{and} \quad \lim_{n \to \infty} a_{2n+1} = -1 \] Since the sequence does not settle on any one value and oscillates indefinitely, the overall conclusion is that the sequence diverges. Thus, the final answer is: DIVERGES