Let \[ a_{n}=\frac{3 n}{6 n+3} \text {. } \] For the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter 'infinity' if it diverges to \( \infty \), '-infinity' if it diverges to \( -\infty \) or 'DNE' otherwise. (a) The series \( \sum_{n=1}^{\infty} \frac{3 n}{6 n+3} \). (b) The sequence \( \left\{\frac{3 n}{6 n+3}\right\} \).

To analyze the sequence and series: For the sequence \( a_n = \frac{3n}{6n+3} \), we can simplify it by dividing the numerator and the denominator by \( n \): \[ a_n = \frac{3}{6 + \frac{3}{n}} \] As \( n \) approaches infinity, \( \frac{3}{n} \) approaches 0, thus: \[ \lim_{n \to \infty} a_n = \frac{3}{6 + 0} = \frac{1}{2} \] So, the sequence converges to \( \frac{1}{2} \). Now, for the series \( \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{3n}{6n+3} \): Here, note that the term \( a_n \) behaves like \( \frac{3}{6} = \frac{1}{2} \) for large \( n \). Thus the series behaves like: \[ \sum_{n=1}^{\infty} \frac{1}{2} \] This series diverges because the terms do not approach zero; instead, they approach \( \frac{1}{2} \). So, summarizing the results: (a) Divergent (DNE) (b) Convergent, limit \( \frac{1}{2} \)