证明:
\[
\lim_{n\rightarrow\infty}\prod_{i=1}^{n+1}\cos\frac{\sqrt{2i-1}}{n}a^2=e^{-\frac{a^4}{2}}.
\]
Remark. 如果取对数, 则等价于
\[
\lim_{n\rightarrow\infty}\sum_{i=1}^{n+1}\ln\cos\frac{\sqrt{2i-1}}{n}a^2=-\frac{a^4}{2}.
\]
1
根据
\[
\ln(1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\cdots
\]
\[
1-\cos x=\frac{1}{2!}x^2-\frac{1}{4!}x^4+\cdots
\]
得
\[
\begin{split}
\ln\cos\frac{\sqrt{2i-1}}{n}a^2&=\cos\frac{\sqrt{2i-1}}{n}a^2-1+o(\cos\frac{\sqrt{2i-1}}{n}a^2-1)^2\\
&=-\frac{1}{2}\frac{2i-1}{n^2}a^4+O((\frac{\sqrt{2i-1}}{n}a^2)^4)+o\Bigl((-\frac{1}{2}\frac{2i-1}{n^2}a^4+O((\frac{\sqrt{2i-1}}{n}a^2)^4))^2\Bigr)\\
&=-\frac{1}{2}\frac{2i-1}{n^2}a^4+O(\frac{(2i-1)^2}{n^4}a^8)+o(\frac{(2i-1)^2}{n^4}a^8)\\
&=-\frac{1}{2}\frac{2i-1}{n^2}a^4+O(\frac{i^2}{n^4})
\end{split}
\]
因此
\[
\begin{split}
\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^{n+1}\ln\cos\frac{\sqrt{2i-1}}{n}a^2}{\sum_{i=1}^{n+1}-\frac{1}{2}\frac{2i-1}{n^2}a^4}&=\lim_{n\rightarrow\infty}\frac{-\frac{a^4}{2}\sum_{i=1}^{n+1}\frac{2i-1}{n^2}+\sum_{i=1}^{n+1}O(\frac{i^2}{n^4})}{-\frac{a^4}{2}\sum_{i=1}^{n+1}\frac{2i-1}{n^2}}\\
&=1+\lim_{n\rightarrow\infty}\frac{O(\frac{1}{n})}{-\frac{a^4}{2}\frac{(n+1)^2}{n^2}}\\
&=1
\end{split}
\]