<s>[INST] 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 $\frac{1}{2}$.
 [/INST] 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 $$\limsup_{n\to \infty}\frac{S_n}{\sqrt n}\le 2\csc(1/2)$$
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