An interesting limit: $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}$ I would like to prove that
$$\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}=0$$
but I am stuck.
I tried to solve it by using Euler-Maclaurin formula, but I could not to.
Euler-Maclaurin formula applied to the function $f(x)=\sin x \sin\sqrt{x}\;\;$ is the following:
$$\sum_{h=1}^n\sin h\sin\sqrt{h}=\int_\limits{0}^n\left[\sin x\sin\sqrt{x}+\left(x-\lfloor x\rfloor\right)\left(\cos x\sin\sqrt{x}+\frac{\sin x\cos\sqrt{x}}{2\sqrt{x}}\right)\right] \, dx$$
but I could not manage to prove that
$$\frac{1}{n}\int_\limits{0}^n\left(x-\lfloor x\rfloor\right)\left(\cos x \sin\sqrt{x} \right) \, dx\rightarrow 0 \text{ as } n\to\infty.$$
Moreover I tried to write the limit as a limit of a Riemann sum, but I did not manage to.
Furthermore I tried to prove the following inequality:
$$\left|\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\cdots+\sin n \sin\sqrt{n} \right|\le\sqrt[4]{n^3}\\\text{for all }\;n\in\mathbb{N},$$
but it was not successful.
I managed to prove that
$$\lim_{n\to\infty}\frac{\sin 1+\sin 2 +\sin 3+\ldots+\sin n}{n}=0$$
and
$$\lim_{n\to\infty}\frac{\sin\sqrt{1}+\sin\sqrt{2}+\sin\sqrt{3}+\cdots+\sin\sqrt{n}}{n}=0.$$
Is it possible to use these last two limits in order to prove that
$$\lim_{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3 \sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}=0\text{ ?}$$
I tried to use Cauchy-Schwartz inequality, but I got $$\lim_{n\to\infty}\frac{\sin^21+\sin^22+\cdots+\sin^2n}{n}$$ and $$\lim_{n\to\infty}\frac{\sin^2\sqrt{1}+\sin^2\sqrt{2}+\cdots+\sin^2\sqrt{n}}{n}$$ and these last two limits are not zero in fact there are both $\frac{1}{2}$.
 A: One can actually say more and show that:
$|\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\ldots+\sin n\sin\sqrt{n}| \le C$ for some universal constant.
Using the sine product formula, it is enough to prove the result for
$$C_1(n)=\cos(1+\sqrt{1})+\cos(2+\sqrt{2})+\cdots+\cos(n+\sqrt{n})$$ and
$$C_2(n)=\cos(1-\sqrt{1})+\cos(2-\sqrt{2})+\cdots+\cos(n-\sqrt{n})$$
and then taking real parts it is enough to show the result for
$$S_{1,2}(n)=\sum_{k=1}^ne^{i(k\pm\sqrt k)}$$
We will show that $|S_{1,2}| \le C$ for a universal constant $C$ so the result will follow and we will do the proof for $S_2$ indicating the estimate changes needed for $S_1$ which are minor.
Let $g(x)=\frac{x-\sqrt x}{2 \pi}, x \ge 1$ and note that $1/(4\pi) \le g'(x) \le 1/(2\pi)$ and this inequality is enough to prove our result (the fact that the lower and upper bounds are constants strictly between $0$ and $1$.
Note also that by omitting a fixed finite number of terms which we can bound trivially the result holds for functions $f(x)$ like $3x+100\sqrt x, -2x+x^{1-1/10000}$ and so on, the crucial part being that $g'(x)=f'(x)/(2\pi)=c+o(1), x \to \infty, c \ne 0, |c| <1$, so $0<c_1<|g'(x)|<c_2<1, x >k$ for constants $c_1,c_2,k$ and for the function $h$ involved in $S_1$ we have $1/(2\pi) \le |h'(x)| \le 3/(4\pi)$
Let $q(n)=g(n+1)-g(n), n \ge 1$ so by the MVT there is $n \le x_n \le n+1, q(n)=g'(x_n)$ In particular $q_n$ increasing since $g'$ does (if $g'$ would be decreasing like for $S_1$ we conjugate and replace $g$ by $-g$) and $1/(4\pi)  \le q(n) \le 1/(2\pi)$
But now the identity:
$$e^{2\pi i g(k)}=1/2(1+i\cot \pi q(k))(e^{2\pi i g(k)}-e^{2\pi i g(k+1)})$$ gives that
$$S_2(n)=\sum_{k=1}^{n}e^{2\pi i g(k)}=\sum_1^{n}1/2(1+i\cot \pi q(k))(e^{2\pi i g(k)}-e^{2\pi i g(k+1)})=$$
$$=i/2\sum_{k=2}^{n-1}e^{2\pi i g(k)}(\cot \pi q(k)-\cot \pi q(k+1))+1/2(1+i\cot \pi q(1))e^{2\pi i g(1)}-(1/2)(1+i\cot \pi q(n))e^{2\pi i g(n+1)}$$
by rearranging the terms and noting that only terms with $g(1), g(n+1)$ appear only once
But now taking absolute values and noting that $\cot \pi q(k)-\cot \pi q(k+1)$ is decreasing since $1/4<\pi q(k) <1/2<\pi, q(k)$ increasing, we get:
$$|S_2(n)| \le 1/2 (\cot \pi q(2)-\cot \pi q(n))+1/2(|\cot \pi q(n)|+|\cot \pi q(1)|+1 \le C_2 $$ where $C_2$ is obtained by using that all the cotangtents above are at most $\cot 1/4$, so one can take $C_2=2\cot 1/4 +1$ for example and clearly we get a similar $C_1$ for $S_1$ so we are done!
A: Let $S_n$ be given by
$$S_n=\sum_{k=1}^n \sin(k)\sin\sqrt{k}\tag1$$
Applying summation by parts to the sum in $(1)$ reveals
$$S_n=\sin(\sqrt {n+1})\sum_{k=1}^{n}\sin(k)-\sum_{k=1}^n \left(\sum_{\ell=1}^k \sin(\ell)\right)\left(\sin(\sqrt {k+1})-\sin(\sqrt{k})\right)\tag 2$$

ESTIMATES:
The sum $\sum_{\ell=1}^k \sin(\ell)$ can be evaluated in closed form which provides the estimate
$$\begin{align}
\left|\sum_{\ell=1}^n \sin(\ell)\right|&=\left|\csc(1/2)\sin(n/2)\sin((n+1)/2)\right|\\\\
\le \csc(1/2)\tag3
\end{align}$$
Moreover, from the Prosthaphaeresis identities, we have the estimate
$$\begin{align}
\left|\sin(\sqrt {k+1})-\sin(\sqrt{k}\right|&=\left|\frac12\cos\left(\frac{\sqrt{k+1}+\sqrt{k}}{2}\right)\sin\left(\frac{\sqrt{k+1}-\sqrt{k}}{2}\right)\right|\\\\
&=\left|2\cos\left(\frac{\sqrt{k+1}+\sqrt{k}}{2}\right)\sin\left(\frac{1}{2(\sqrt{k+1}+\sqrt{k})}\right)\right|\\\\
&\le \frac{1}{\sqrt{k}}\tag4
\end{align}$$

Using the estimates in $(3)$ and $(4)$ in $(2)$, we find that
$$\begin{align}
|S_n|&\le \csc(1/2)\left(1+\sum_{k=1}^n\frac1{\sqrt k}\right)\\\\
&\le \csc(1/2)(1+2\sqrt n)\tag5
\end{align}$$

Finally, using the estimate in $(5)$ we have
$$\left|\frac{S_n}{n}\right|\le \frac{\csc(1/2)(1+2\sqrt n)}{n}$$
whence application of the squeeze theorem recovers the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\sum_{k=1}^n \sin(k)\sin(\sqrt k)}{n}=0}$$

NOTE:  We have tacitly found that $$\limsup_{n\to \infty}\frac{S_n}{\sqrt n}\le 2\csc(1/2)$$
