35. \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) 37. \( \sum_{n=1}^{\infty} \ln \frac{1}{n} \)

The series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) can be approximated using the Taylor series expansion for sin(x) around 0, which states that \(\sin x \approx x\). Thus, \(\sin \frac{1}{n} \approx \frac{1}{n}\) for large n. The series behaves like \( \sum_{n=1}^{\infty} \frac{1}{n} \), which is a well-known divergent harmonic series. For the series \( \sum_{n=1}^{\infty} \ln \frac{1}{n} \), we can rewrite it as \( -\sum_{n=1}^{\infty} \ln n \). Similar to the first series, we can use the integral test or properties of p-series to show divergence. The logarithmic function grows slowly but is unbounded as n increases, so the series does not converge and diverges to infinity.