無限級数その８／適用その４

［適用その４］［定理05］から，
$$\sum_{n=1}^{∞}( \frac{1}{2n-1}- \frac{1}{2n} )= \log2． \cdots (121) $$
$$\sum_{n=1}^{∞}( \frac{1}{3n-1}- \frac{1}{3n} )= -\frac{\sqrt{3}\pi}{18} +\frac{\log3}{2}． \cdots (131) $$
$$\sum_{n=1}^{∞}( \frac{1}{3n-2}- \frac{1}{3n} )= \frac{\sqrt{3}\pi}{18} +\frac{\log3}{2}． \cdots (132) $$
$$\sum_{n=1}^{∞}( \frac{1}{4n-1}- \frac{1}{4n} )= -\frac{\pi}{8} +\frac{3\log2}{4}． \cdots (141) $$
$$\sum_{n=1}^{∞}( \frac{1}{4n-3}- \frac{1}{4n} )= \frac{\pi}{8} +\frac{3\log2}{4}． \cdots (143) $$