無限級数その８／適用その４

［適用その４］［定理05］から，
$$\sum _{n=1}^{\infty }(\frac{1}{2n-1}-\frac{1}{2n})=\log 2．\cdots (121)$$
$$\sum _{n=1}^{\infty }(\frac{1}{3n-1}-\frac{1}{3n})=-\frac{\sqrt{3}\pi }{18}+\frac{\log 3}{2}．\cdots (131)$$
$$\sum _{n=1}^{\infty }(\frac{1}{3n-2}-\frac{1}{3n})=\frac{\sqrt{3}\pi }{18}+\frac{\log 3}{2}．\cdots (132)$$
$$\sum _{n=1}^{\infty }(\frac{1}{4n-1}-\frac{1}{4n})=-\frac{\pi }{8}+\frac{3\log 2}{4}．\cdots (141)$$
$$\sum _{n=1}^{\infty }(\frac{1}{4n-3}-\frac{1}{4n})=\frac{\pi }{8}+\frac{3\log 2}{4}．\cdots (143)$$