Find a limit as product of cos $$\lim\limits_{n\to\infty}\cos\frac{1}{n\sqrt{n}}\cos\frac{2}{n\sqrt{n}}\cdots\cos\frac{n}{n\sqrt{n}}$$
It is not hard to prove that the limit exists, but is that possible to calculate the limit? Suggestions are welcome!
 A: The infinite product is a bit difficult to work with, so to make things easier we compute the log of the product instead. Our limit becomes
$$\lim_{n\to\infty}\sum^n_{i=1}\ln\cos\frac{i}{n^{\frac{3}{2}}}$$
We use a Taylor series to compute $\ln\cos x$ around $x=0$. We know $\cos x=1-\frac{x^2}{2}+O\left(x^4\right)$ and $\ln(1+x)=x+O\left(x^2\right)$. Hence,
$$\ln\cos x=\ln\left(1-\frac{x^2}{2}+O\left(x^4\right)\right)=-\frac{x^2}{2}+O\left(x^4\right)+O\left(\left(-\frac{x^2}{2}+O\left(x^4\right)\right)^2\right)$$
The last asymptotic is equal to $O\left(x^4\right)$, so our final expression becomes
$$\ln\cos x=-\frac{x^2}{2}+O\left(x^4\right)$$
Applying this in our limit we get:
$$\lim_{n\to\infty}\sum^n_{i=1}-\frac{i^2}{2n^3}+O\left(\frac{i^4}{n^6}\right)=\lim_{n\to\infty}-\sum^n_{i=1}\frac{i^2}{2n^3}+\sum^n_{i=1}O\left(\frac{i^4}{n^6}\right)$$
Now, $\sum^n_{i=1}O\left(i^k\right)=O\left(n^{k+1}\right)$ and $\sum^n_{i=1}i^2=\frac{2n^3+3n^2+n}{6}$ in particular, so our limit becomes
$$\lim_{n\to\infty}-\frac{2n^3+3n^2+n}{12n^3}+O\left(\frac{n^5}{n^6}\right)=-\frac{1}{6}+O\left(n^{-1}\right)$$
Now $O\left(n^{-1}\right)\sim0$ for large $n$, so we conclude
$$\lim_{n\to\infty}\sum^n_{i=1}\ln\cos\frac{i}{n^{\frac{3}{2}}}=-\frac{1}{6}$$
Or equivalently
$$\lim_{n\to\infty}\prod^n_{i=1}\cos\frac{i}{n^{\frac{3}{2}}}=e^{-\frac{1}{6}}$$
The LHS is your limit.
A: $$P_n=\prod _{i=1}^{n } \cos \left(\frac{i}{n\sqrt{n} }\right)\implies \log(P_n)=\sum _{i=1}^{n } \log \left(\cos \left(\frac{i}{n\sqrt{n} }\right)\right)$$ Using the usual Taylor series of $\cos(x)$ around $x=0$
$$\log(\cos(x))=\sum _{k=1}^{\infty } (-1)^k \,\frac{2^{2 k-3} (E_{2 k-1}(1)-E_{2 k-1}(0)) }{k (2 k-1)!}x^{2 k}$$  where appear Euler numbers and Euler polynomials.
Making $x=\frac{i}{n\sqrt{n} }$, summing to some order
$$\log(P_n)=-\frac{1}{6}-\frac{4}{15 n}-\frac{323}{2520 n^2}-\frac{899}{22680
   n^3}+O\left(\frac{1}{n^4}\right)$$
$$P_n=e^{\log(P_n)}=e^{-\frac{1}6}\Bigg[1-\frac{4}{15 n}-\frac{389}{4200 n^2}-\frac{181}{21000
   n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$ So, the limit and a good approximation of the partial product. For example, this truncated expansion gives
$$P_{10}=\frac{20420369}{21000000} \,e^{-\frac{1}6}=0.8231175797$$ while the exact value is $0.8231175865$.
Now, if you want to make it looking smart, using Pochhammer symbols, we have
$$P_n=\frac 1{2^{n+1}} \exp\Bigg[  \frac{n+1}{2 \sqrt{n}}i\Bigg]\left(-1;e^{-\frac{2 }{n\sqrt{n}}i}\right){}_{n+1}$$
