A thinking problem of limit from my teacher. Please find the limit:$$\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) +  \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}} \right)} \right)} \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right].$$
 A: Step 1. Since for small $x$ we have $\frac{1}{\sqrt{1+x}}=1-\frac{x}{2}+\frac{3x^2}{8}+O(x^3),$
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}=\frac{1}{n}\sum_{k=1}^{n}\left(1-\frac{k}{2n^2}+\frac{3k^2}{8n^4}\right)+O\left(\frac{1}{n^3}\right)$$
hence:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=1-\frac{1}{4n}-\frac{1}{8n^2}+O\left(\frac{1}{n^3}\right).$$
Step 2.
$$\sum_{k=1}^{n}\frac{1}{\left(\sqrt{n^2+k}\right)^3}=\frac{1}{n^2}+O\left(\frac{1}{n^3}\right).$$
Step 3. Since for small $x$ we have $\sin x = x-\frac{x^3}{6}+O(x^5)$,
$$\sum_{k=1}^{n}\sin\frac{\pi}{\sqrt{n^2+k}}=\pi-\frac{\pi}{4n}-\frac{\pi}{8n^2}-\frac{\pi^3}{6n^2}+O\left(\frac{1}{n^3}\right).$$
Step 4. Since for small $x$ we have $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$,
$$ n\log\left(\frac{1}{\pi}\sum_{k=1}^{n}\sin\frac{\pi}{\sqrt{n^2+k}}\right)=-\frac{1}{4}-\left(\frac{5}{32}+\frac{\pi^2}{6}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right).$$
Step 5. Exponentiating, it follows that the value of the limit is:

$$ L = -\frac{15+16\,\pi^2}{96\, e^{1/4}}.$$

A: Since $n$ is supposed to be large, then the argument of the sine is small. So, start with $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ Now, replace $x$ by $\frac{\pi }{\sqrt{k+n^2}}$ and then $$\frac{1 }{\sqrt{k+n^2}}=\frac{1}{n}-\frac{k}{2 n^3}+\frac{3 k^2}{8
   n^5}+O\left(\left(\frac{1}{n}\right)^6\right)$$ So, $$\sin \left(\frac{\pi }{\sqrt{k+n^2}}\right)\approx\frac{\pi }{n}-\frac{\frac{\pi  k}{2}+\frac{\pi ^3}{6}}{n^3}$$ Now, sum over $k$ from $1$ to $n$ and the sum of sines is $$\pi-\frac{\pi }{4 n}-\frac{\pi  \left(3+2 \pi ^2\right)}{12 n^2}\approx \pi-\frac{\pi }{4 n}$$ So, what is inside brackets write $$\left(1-\frac{1}{4 n}\right)^n-\frac{1}{\sqrt[4]{e}}$$
I am sure that you recognize the first term and that you can take from here.
