求证：
\[ \lim_{n\to\infty}n\left[\left(\frac{1}{\pi}\left(\sin\left(\frac{\pi}{\sqrt{n^2+1}}\right)+\sin\left(\frac{\pi}{\sqrt{n^2+2}}\right)+\cdots+\sin\left(\frac{\pi}{\sqrt{n^2+n}}\right) \right)\right)^{n}-\frac{1}{\sqrt[4]{e}}\right]=-\frac{1}{\sqrt[4]{e}}\left(\frac{15}{96}+\frac{\pi^2}{6} \right)   \]

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证明
记
\[ I=n\left[\left(\frac{1}{\pi}\left(\sin\left(\frac{\pi}{\sqrt{n^2+1}}\right)+\sin\left(\frac{\pi}{\sqrt{n^2+2}}\right)+\cdots+\sin\left(\frac{\pi}{\sqrt{n^2+n}}\right) \right)\right)^{n}-\frac{1}{\sqrt[4]{e}}\right]\]
则
\[ I=\frac{n}{\sqrt[4]{e}}\left(\exp\left(n\ln\frac{\sin\frac{\pi}{\sqrt{n^2+1}}+\sin\frac{\pi}{\sqrt{n^2+2}}+\cdots+\sin\frac{\pi}{\sqrt{n^2+n}}}{\pi}+\frac{1}{4}\right)-1 \right)\]
注意到
\[ \sin\frac{\pi}{\sqrt{n^2+k}}=\frac{\pi}{\sqrt{n^2+k}}-\frac{1}{6}\left(\frac{\pi}{\sqrt{n^2+k}}\right)+o\left(\frac{1}{n^3}\right)\qquad (n\to+\infty) \]
所以
\[ \frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]+o\left(\frac{1}{n^2}\right)\qquad (n\to+\infty) \]
而
\begin{align*}
 \sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}&=\frac{1}{n}\sum_{k=1}^{n}\left(1+\frac{k}{n^2}\right)^{-\frac{1}{2}}\\
&=\frac{1}{n}\sum_{k=1}^{n}\left(1-\frac{k}{2n^2}+\frac{3}{8}\left(\frac{k}{n^2}\right)^{2}+o\left(\frac{1}{n^2}\right)\right)\\
&=1-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}+o\left(\frac{1}{n^2}\right)
\end{align*}
所以
\[ \frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)=1-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]+o\left(\frac{1}{n^2}\right) \]
\begin{align*}
&\ln\left[\frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right]\\
&=-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]-\frac{1}{2}\left[ \frac{(n+1)}{4n^2}-\frac{(n+1)(2n+1)}{16n^4}+\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]\right]^{2}+o\left(\frac{1}{n^2}\right)\\
&=-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]-\frac{1}{2}\left[\frac{n+1}{4n^2}+o\left(\frac{1}{n}\right)\right]^2+o\left(\frac{1}{n^2}\right)\qquad (n\to+\infty)
\end{align*}
\begin{align*}
&n\ln\left[\frac{1}{\pi}\sum_{k=1}^{n}\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right]+\frac{1}{4}\\
&=\frac{1}{4}+n\left[-\frac{(n+1)}{4n^2}+\frac{(n+1)(2n+1)}{16n^4}-\frac{1}{6}\left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]-\frac{1}{2}\left[\frac{n+1}{4n^2}+o\left(\frac{1}{n}\right)\right]^2+o\left(\frac{1}{n^2}\right)\right]\\
&=-\frac{15}{96n}-\frac{\pi^2}{6n}+o\left(\frac{1}{n}\right)\quad  (n\to+\infty)
\end{align*}
这里得注意到事实
\[ \left[\sum_{k=1}^{n}\frac{\pi^2}{\sqrt{(n^2+k)^3}}\right]\sim \frac{\pi^2}{n^2} \]

所以就有
\[ \lim_{n\to\infty}I=\lim_{n\to\infty}\frac{n}{\sqrt[4]{e}}\left(e^{-\frac{15}{96n}-\frac{\pi^2}{6n}+o\left(\frac{1}{n}\right)}-1\right)= -\frac{1}{\sqrt[4]{e}}\left(\frac{15}{96}+\frac{\pi^2}{6} \right) \]