1. **State the problem:** We want to evaluate the infinite series $$\sum_{n=1}^{\infty} \left( \ln \sqrt{n+1} \mp \ln \sqrt{n} \right).$$ 2. **Rewrite the terms:** Recall that $\ln \sqrt{x} = \frac{1}{2} \ln x$. So each term becomes $$\ln \sqrt{n+1} \mp \ln \sqrt{n} = \frac{1}{2} \ln (n+1) \mp \frac{1}{2} \ln n = \frac{1}{2} \left( \ln (n+1) \mp \ln n \right).$$ 3. **Consider the two cases for the sign:** - If the sign is minus ($-$), the term is $$\frac{1}{2} \left( \ln (n+1) - \ln n \right) = \frac{1}{2} \ln \frac{n+1}{n} = \frac{1}{2} \ln \left(1 + \frac{1}{n} \right).$$ - If the sign is plus ($+$), the term is $$\frac{1}{2} \left( \ln (n+1) + \ln n \right) = \frac{1}{2} \ln \left( n(n+1) \right).$$ 4. **Evaluate the series for the minus case (telescoping):** The series is $$\sum_{n=1}^\infty \frac{1}{2} \ln \frac{n+1}{n} = \frac{1}{2} \sum_{n=1}^\infty \ln \frac{n+1}{n}.$$ Write partial sums: $$S_N = \frac{1}{2} \sum_{n=1}^N \ln \frac{n+1}{n} = \frac{1}{2} \ln \prod_{n=1}^N \frac{n+1}{n} = \frac{1}{2} \ln \frac{2}{1} \cdot \frac{3}{2} \cdots \frac{N+1}{N}.$$ Notice the telescoping product: $$\frac{2}{1} \cdot \frac{3}{2} \cdots \frac{N+1}{N} = \frac{N+1}{1} = N+1.$$ So $$S_N = \frac{1}{2} \ln (N+1).$$ 5. **Take the limit as $N \to \infty$:** $$\lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{1}{2} \ln (N+1) = \infty,$$ so the series diverges to infinity. 6. **Evaluate the series for the plus case:** The terms are $$\frac{1}{2} \ln \left( n(n+1) \right) = \frac{1}{2} \left( \ln n + \ln (n+1) \right).$$ The series is $$\sum_{n=1}^\infty \frac{1}{2} \left( \ln n + \ln (n+1) \right) = \frac{1}{2} \sum_{n=1}^\infty \ln n + \frac{1}{2} \sum_{n=1}^\infty \ln (n+1).$$ Both sums diverge to infinity, so the series diverges. **Final answer:** - For the minus sign, the series diverges to infinity. - For the plus sign, the series also diverges. Hence, the series does not converge for either sign.