How find this limits with hardly form? 
show that:
$$\lim_{n\to\infty}n\left[\left(\dfrac{1}{\pi}\left(\sin{\left(\dfrac{\pi}{\sqrt{n^2+1}}\right)}+
\sin{\left(\dfrac{\pi}{\sqrt{n^2+2}}\right)}+\cdots+\sin{\left(\dfrac{\pi}{\sqrt{n^2+n}}\right)}
\right)\right)^n-\dfrac{1}{\sqrt[4]{e}}\right]=-\dfrac{1}{\sqrt[4]{e}}\left(\dfrac{31}{96}+\dfrac{\pi^2}{6}\right)$$

I only solve this :$$\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\dfrac{1}{\sqrt[4]{e}}\tag{1}$$
use
$$\sin{x}\approx x,\Longrightarrow \dfrac{1}{\pi}\sin{(\dfrac{\pi}{\sqrt{n^2+i}})}=\dfrac{1}{\sqrt{n^2+i}}+o(\dfrac{1}{\sqrt{n^2+i}})\to \dfrac{1}{n}\left(1+\dfrac{i}{n^2}\right)^{-\frac{1}{2}},n\to\infty$$
and note
$$(1+x)^{-1/2}=1-\dfrac{1}{2}x+o(x)$$
so
$$\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1}{n}\sum_{i=1}^{n}\left(1-\dfrac{i}{2n^2}+o(i/n^2)\right)\right)^n=e^{-\frac{1}{4}}$$
But I can't solve my problem,Thank you
 A: You seem to be on the right track and I guess we can try to get more reasonable estimates using further terms of the expansion. First we need to use $\sin x \approx x - \dfrac{x^{3}}{6}$ to get $$\begin{aligned}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx  \frac{1}{n}\left(1 + \frac{i}{n^{2}}\right)^{-1/2} - \frac{\pi^{2}}{6n^{3}}\left(1 + \frac{i}{n^{2}}\right)^{-3/2}\\
&\approx \frac{1}{n}\left(1 - \frac{i}{2n^{2}} + \frac{3i^{2}}{8n^{4}}\right) - \frac{\pi^{2}}{6n^{3}}\left(1 - \frac{3i}{2n^{2}} + \frac{15i^{2}}{8n^{4}}\right)\\
&\approx \frac{1}{n}\left(1 - \frac{3i + \pi^{2}}{6n^{2}} + \frac{9i^{2} + 6i\pi^{2}}{24n^{4}}\right)\end{aligned}$$ and then summing from $i = 1$ to $i = n$ we get $$\begin{aligned}\sum_{i = 1}^{n}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx \frac{1}{n}\left(n - \dfrac{\dfrac{3n(n + 1)}{2} + n\pi^{2}}{6n^{2}} + \dfrac{\dfrac{3n(n + 1)(2n + 1)}{2} + 3n(n + 1)\pi^{2}}{24n^{4}}\right)\\
&= \left(1 - \frac{3(n + 1) + 2\pi^{2}}{12n^{2}} + \frac{(n + 1)(2n + 1) + 2(n + 1)\pi^{2}}{16n^{4}}\right)\\
&\approx \left(1 - \frac{1}{4n} - \frac{2\pi^{2} + 3}{12n^{2}} + \frac{1}{8n^{2}}\right)\\
&= \left(1 - \frac{1}{4n} - \frac{4\pi^{2} + 3}{24n^{2}}\right)\\\end{aligned}$$ and therefore using logs we get $$\begin{aligned}\left(\sum_{i = 1}^{n}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right)\right)^{n} &\approx \exp\left(n\log\left(1 - \frac{1}{4n} - \frac{4\pi^{2} + 3}{24n^{2}}\right)\right)\\
&\approx\exp\left(-n\left(\frac{1}{4n} + \frac{\pi^{2}}{6n^{2}} + \frac{5}{32n^{2}}\right)\right)\\
&= \exp\left(-\frac{1}{4} - \frac{\pi^{2}}{6n} - \frac{5}{32n}\right)\\ \end{aligned}$$ We can now see that the desired limit is equal $$\begin{aligned}L &= \lim_{n \to\infty}n\left(\exp\left(-\frac{1}{4} - \frac{\pi^{2}}{6n} - \frac{5}{32n}\right) - \exp(-1/4)\right)\\
&= e^{-1/4}\lim_{n \to\infty}n\left(\exp\left(- \frac{\pi^{2}}{6n} - \frac{5}{32n}\right) - 1\right) = -\frac{1}{\sqrt[4]{e}}\left(\frac{\pi^{2}}{6} + \frac{5}{32}\right)\end{aligned}$$ There might be some calculation mistake in above or there is some typo in the question.
