Evaluate the infinite sum S=∑n=0∞(−1)n3n+1S = \sum_{n=0}^{\infty} \frac{(-1)^n}{3n+1}S=∑n=0∞3n+1(−1)n.
13ln2+π33\frac{1}{3} \ln 2 + \frac{\pi}{3\sqrt{3}}31ln2+33π
13ln2+π63\frac{1}{3} \ln 2 + \frac{\pi}{6\sqrt{3}}31ln2+63π
23ln2+π33\frac{2}{3} \ln 2 + \frac{\pi}{3\sqrt{3}}32ln2+33π
16ln2+π23\frac{1}{6} \ln 2 + \frac{\pi}{2\sqrt{3}}61ln2+23π