置 y_n:=2\cos\frac{a}{2^n}-1, 则有
y_{n}\cos\frac{a}{2^{n+1}}=\left(4\cos^2\frac{a}{2^{n+1}}-3\right)\cos\frac{a}{2^{n+1}}=\cos\frac{3a}{2^{n+1}},
于是
y_n=\frac{\cos\frac{3a}{2^{n+1}}}{\cos\frac{a}{2^{n+1}}},
所以
\prod_{k=1}^ny_k=\prod_{k=1}^n\frac{\cos\frac{3a}{2^{k+1}}}{\cos\frac{a}{2^{k+1}}}=\prod_{k=1}^n\frac{\frac{\sin\frac{3a}{2^k}}{2\sin\frac{3a}{2^{k+1}}}}{\frac{\sin\frac{a}{2^k}}{2\sin\frac{a}{2^{k+1}}}}=\frac{\sin\frac{3a}{2}}{\sin\frac{a}{2}}\cdot\frac{\sin\frac{a}{2^{n+1}}}{\sin\frac{3a}{2^{n+1}}},
于是
\prod_{k=1}^{\infty}y_k=\lim_{n \to \infty}\left(\frac{\sin\frac{3a}{2}}{\sin\frac{a}{2}}\cdot\frac{\sin\frac{a}{2^{n+1}}}{\sin\frac{3a}{2^{n+1}}}\right)=\lim_{n \to \infty}\left(\frac{\sin\frac{3a}{2}}{\sin\frac{a}{2}}\cdot\frac{\frac{a}{2^{n+1}}}{\frac{3a}{2^{n+1}}}\right)=\frac{2\cos a+1}{3},
所以
\sum_{n=1}^{\infty}\ln y_n=\ln\prod_{n=1}^{\infty} y_n=\ln\left(\frac{2\cos a+1}{3}\right).
