ゴールドバッハ・オイラーの定理
1 2 2 − 1 + 1 2 3 − 1 + 1 3 2 − 1 + 1 4 2 − 1 + 1 5 2 − 1 + . . . = 1 3 + 1 7 + 1 8 + 1 15 + 1 24 + . . . < 1 3 + 1 4 + 1 8 + 1 9 + 1 16 + . . . = 1 3 + ∑ m = 2 ∞ ∑ k = 2 ∞ 1 m k = 1 3 + ∑ m = 2 ∞ 1 m 2 ∑ k = 0 ∞ 1 m k = 1 3 + ∑ m = 2 ∞ 1 m 2 m m − 1 = 1 3 + ∑ m = 2 ∞ ( 1 m − 1 − 1 m ) = 1 3 + 1 = 4 3 {\displaystyle {\begin{aligned}&\quad {\frac {1}{2^{2}-1}}+{\frac {1}{2^{3}-1}}+{\frac {1}{3^{2}-1}}+{\frac {1}{4^{2}-1}}+{\frac {1}{5^{2}-1}}+...\\&={\frac {1}{3}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{15}}+{\frac {1}{24}}+...\\&<{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{9}}+{\frac {1}{16}}+...\\&={\frac {1}{3}}+\sum _{m=2}^{\infty }\sum _{k=2}^{\infty }{\frac {1}{m^{k}}}\\&={\frac {1}{3}}+\sum _{m=2}^{\infty }{\frac {1}{m^{2}}}\sum _{k=0}^{\infty }{\frac {1}{m^{k}}}\\&={\frac {1}{3}}+\sum _{m=2}^{\infty }{\frac {1}{m^{2}}}{\frac {m}{m-1}}={\frac {1}{3}}+\sum _{m=2}^{\infty }\left({\frac {1}{m-1}}-{\frac {1}{m}}\right)={\frac {1}{3}}+1={\frac {4}{3}}\\\end{aligned}}}