Finding the limit of $\frac{\sqrt{n^{3n}}}{n!} \prod_{1\leq k\leq n} \sin\left(\frac{k}{n^{3/2}}\right)$ as $n\to\infty$ I found the problem 1.4 in chapter II of "Selected Problems in Real Analysis" really challenging:

Compute the limit of the sequence ${x_n}$, where $$x_n= \frac{\sqrt{n^{3n}}}{n!} \prod_{1\leq k\leq n} \sin\left(\frac{k}{n^{\frac32 }}\right)$$

Has anyone done this problem before, I really can't understand the solution in this book.

The suggested solution in the book:
Write $x_n$ in the form of $$
x_{n}=\prod_{1 \leq k \leq n} \frac{\sin \left(k / n^{3 / 2}\right)}{k / n^{3 / 2}}
$$
and study $\ln x_n$ using Taylor's formula.
 A: For very large $n$, $k / n^{3/2} \leq 1 / \sqrt{n}$ since $k \leq n$. Using that $\sin(x) \approx x$ for $x$ smalls, you have that the product can be approximated by $\prod k / n^{3/2}$. So you expect the product to be approximately $n! / (n^{3n/2})$. 
However, using only that 
$$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
is not enough, as evidence by the following example. 
Example
Let $f(x) = x + x^2$. Then $\lim_{x \to 0} f(x)/x = 1$. However, the limit
$$ \lim_{n \to \infty} \frac{n^{3n/2}}{n!} \prod_{1\leq k\leq n} f( \frac{k}{n^{3/2}}) $$
actually diverges. 
Proof:
$$ \frac{n^{3n/2}}{n!} \prod_{1\leq k\leq n} f( \frac{k}{n^{3/2}}) = \prod_{1 \leq k \leq n} (1 + \frac{k}{n^{3/2}}) = 1 + \sum_{k = 1}^n \frac{k}{n^{3/2}} + \cdots $$
where the omitted terms are all positive. 
This tells us that 
$$ \frac{n^{3n/2}}{n!} \prod_{1\leq k\leq n} f( \frac{k}{n^{3/2}}) \geq 1 + \sum_{k = 1}^n \frac{k}{n^{3/2}} = 1 + \frac{n(n+1)}{2 n^{3/2}} $$
this grows like $\sqrt{n}$ as $n\to \infty$, showing that the original sequence diverges. 

So it is crucially important that $\sin(x)$ does not have a quadratic contribution. 
In fact, as we will see below, even the cubic term in the Taylor expansion makes a non-zero contribution, and so the naive expectation that the limit converges to 1 is in fact false!
What we will use is that 
$$ |\sin(x) - x + \frac{1}{6} x^3| \leq \frac{1}{120} x^5$$
a consequence of Taylor's theorem. This implies
$$  \prod_{k = 1}^n (1 - \frac16 \frac{k^2}{n^{3}}) \leq x_n \leq \prod_{k = 1}^n (1 - \frac{1}{6} \frac{k^2}{n^{3}}+ \frac{1}{120} \frac{k^4}{n^6}) $$
One can fairly simply estimate
$$ \prod_{k = 1}^n (1 - \frac{1}{6} \frac{k^2}{n^{3}}+ \frac{1}{120} \frac{k^4}{n^6}) - \prod_{k = 1}^n (1 - \frac16 \frac{k^2}{n^{3}}) = O(1/n) $$
and so the limit should be 
$$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \prod_{k = 1}^n (1 - \frac16 \frac{k^2}{n^{3}}) $$
To compute this final limit, we take the log of the expression 
$$ \ln \prod_{k = 1}^n (1 - \frac16 \frac{k^2}{n^{3}}) = \sum_{k = 1}^n \ln(1 - \frac16 \frac{k^2}{n^{3}}) =  \sum_{k = 1}^n (-\frac16 \frac{k^2}{n^3} + O(\frac{1}{n^2})) $$ 
which implies
$$ \ln \prod_{k = 1}^n (1 - \frac16 \frac{k^2}{n^{3}}) = - \frac{2 n^3 + 3 n^2 + n}{36n^3} + O(1/n) $$
so 
$$ \lim_{n \to \infty} \ln \prod_{k = 1}^n (1 - \frac16 \frac{k^2}{n^{3}})  = - 1/18 $$
so the limit of the original sum converges to (provided I didn't do any boneheaded algebraic mistakes)
$$ \lim_{n \to \infty} x_n = e^{-1/18} < 1$$
A: Consider the following limit problem

$$ \lim_{n\to\infty} \frac{n^{3n/2}}{n!} \prod_{k=1}^{n} \left( \frac{k}{n^{3/2}} e^{-ck^2/n^3} \right) = ? $$

By direct computation,
$$ \frac{n^{3n/2}}{n!} \prod_{k=1}^{n} \left( \frac{k}{n^{3/2}} e^{-ck^2/n^3} \right)
= \exp\left\{ - c \sum_{k=1}^{n} \frac{k^2}{n^3} \right\}
\xrightarrow[n\to\infty]{} \exp\left\{ - c \int_{0}^{1} x^2 \, dx \right\}
= e^{-c/3}. $$
This obvious computation is in fact the key step towards OP's problem. Indeed, notice that
$$ \sin\left(\frac{k}{n^{3/2}}\right)
= \frac{k}{n^{3/2}} \exp\left\{ \log \left( \frac{\sin(k/n^{3/2})}{k/n^{3/2}} \right) \right\}. $$
Now by Taylor's approximation, we easily find that, as $x \to 0$,
$$ \log \left( \frac{\sin x}{x} \right) = \log \left( 1 - \frac{x^2}{6} + \mathcal{O}(x^4) \right) = -\frac{x^2}{6} + \mathcal{O}(x^4). $$
So it follows that
$$ \exp\left\{ \log \left( \frac{\sin(k/n^{3/2})}{k/n^{3/2}} \right) \right\}
= \exp\left\{ -\frac{k^2}{6n^3} + \mathcal{O}\left( \frac{k^4}{n^6} \right) \right\} $$
Mimicking the previous computation, this tells that

$$ \frac{n^{3n/2}}{n!} \prod_{k=1}^{n} \sin\left(\frac{k}{n^{3/2}}\right)
= \exp\left\{ - \sum_{k=1}^{n} \left( \frac{k^2}{6n^3} + \mathcal{O}\left( \frac{k^4}{n^6} \right) \right) \right\}
\xrightarrow[n\to\infty]{} e^{-1/18}. $$

