米林级数

定理(Millin, Good) $$ \sum_{n=0}^\infty\frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}. $$

引理1 $$ F_{n+1}F_{n-1}-F_{n}^2=(-1)^n. $$

引理2 $$ F_{m+n}=F_{m-1} F_n+F_m F_{n+1}. $$

引理3 $$ F_{m+n-1}=F_{m} F_n+F_{m-1}F_{n-1}. $$

引理4 $$ \lim_{n\to\infty}\frac{F_{n-1}}{F_{n}}=\frac{\sqrt{5}-1}{2}. $$

米林级数的证明 先证 $$ \sum_{n=0}^k\frac{1}{F_{2^n}}=3-\frac{F_{2^k-1}}{F_{2^k}}, $$ $k=1$ 时显然成立, 而 \begin{align*} 3-\frac{F_{2^k-1}}{F_{2^k}}+\frac{1}{F_{2^{k+1}}}=&3-\left(\frac{F_{2^k-1}}{F_{2^k}}-\frac{1}{F_{2^{k+1}}}\right)\\ =&3-\frac{\frac{F_{2^{k}-1}F_{2^{k+1}}}{F_{2^{k}}}-1}{F_{2^{k+1}}}\\ =&3-\frac{F_{2^{k}-1}(F_{2^{k}-1}+F_{2^{k}+1})-1}{F_{2^{k+1}}}\\ =&3-\frac{F_{2^{k}-1}^2+F_{2^{k}-1}F_{2^{k}+1}-1}{F_{2^{k+1}}}\\ =&3-\frac{F_{2^{k}-1}^2+F_{2^{k}}^2}{F_{2^{k+1}}}\\ =&3-\frac{F_{2^{k+1}-1}}{F_{2^{k+1}}} \end{align*} 由归纳法知 $$ \sum_{n=0}^k\frac{1}{F_{2^n}}=3-\frac{F_{2^k-1}}{F_{2^k}}, $$ 令 $k\to\infty$, 即得 $$ \sum_{n=0}^\infty\frac{1}{F_{2^n}}=3-\frac{\sqrt{5}-1}{2}=\frac{7-\sqrt{5}}{2}. $$