TITLE: An interesting limit: $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}$
QUESTION [18 upvotes]: 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 $\frac{1}{2}$.

REPLY [2 votes]: Property 1:
If $\;\left\{a_n\right\}_{n\in\mathbb{N}}\;$ is a sequence of real numbers such that $\;\left\{a_n-a_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonic and there exists $\;k\in\mathbb{Z}\;$ for which $\;2\pi k<a_n-a_{n-1}<2\pi+2\pi k\;\;\;\;\forall n\in\mathbb{N}\setminus\{1\}\;,\;\;\;$ then $$\left|\sum_\limits{h=1}^n \cos a_h\right|\le\frac{1}{2}\left[\;\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)-\cot\left(\frac{a_2-a_1}{2}\right)\right|+\\+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+|\sin a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+\\+|\cos a_1|+1\;\right]$$ for all $\;n\in\mathbb{N}\setminus\{1\}.$
Proof:
By applying Prosthaphaeresis identities, we get that
$\cos a_h+\cos a_{h+1}=2\cos\left(\frac{a_{h+1}+a_h}{2}\right)\cos\left(\frac{a_{h+1}-a_h}{2}\right)=\\=2\cos\left(\frac{a_{h+1}+a_h}{2}\right)\sin\left(\frac{a_{h+1}-a_h}{2}\right)\cot\left(\frac{a_{h+1}-a_h}{2}\right)=\\=\left(\sin a_{h+1}-\sin a_h\right)\cot\left(\frac{a_{h+1}-a_h}{2}\right)\;,\;\;\text{ for all }h\in\mathbb{N}.$
Moreover,
$2\sum_\limits{h=1}^n\cos a_h=\sum_\limits{h=1}^{n-1}\left(\cos a_h +\cos a_{h+1}\right)+\cos a_1+\cos a_n=\\=\sum_\limits{h=1}^{n-1}\left(\sin a_{h+1}-\sin a_h\right)\cot\left(\frac{a_{h+1}-a_h}{2}\right)+\cos a_1+\cos a_n =\\=\sum_\limits{h=1}^{n-1}\sin a_{h+1}\cot\left(\frac{a_{h+1}-a_h}{2}\right)-\sum_\limits{h=1}^{n-1}\sin a_h\cot\left(\frac{a_{h+1}-a_h}{2}\right)+\\+\cos a_1+\cos a_n=\\=\sum_\limits{h=2}^{n}\sin a_h\cot\left(\frac{a_h-a_{h-1}}{2}\right)-\sum_\limits{h=1}^{n-1}\sin a_h\cot\left(\frac{a_{h+1}-a_h}{2}\right)+\\+\cos a_1+\cos a_n =\\=\sum_\limits{h=2}^{n-1}\sin a_h\left[\cot\left(\frac{a_h-a_{h-1}}{2}\right)-\cot\left(\frac{a_{h+1}-a_h}{2}\right)\right]+\\+\sin a_n\cot\left(\frac{a_n-a_{n-1}}{2}\right)-\sin a_1\cot\left(\frac{a_2-a_1}{2}\right)+\cos a_1+\cos a_n\;,\\\text{ for all }\;n\in\mathbb{N}\setminus\{1\}.$
Since the function $\;\cot\;$ is monotonic on $\;\left]\pi k,\pi+\pi k\right[\;$ and $\;\left\{a_n-a_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is a monotonic sequence such that $\;2\pi k<a_n-a_{n-1}<2\pi+2\pi k\;\;\;\;\forall n\in\mathbb{N}\setminus\{1\}\;,\;$ then the sequence $\;\left\{\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonic too.
So by taking absolute values and by noting that the sequence $\;\left\{\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonic, we get that
$2\left|\sum_\limits{h=1}^n\cos a_h\right|\le\sum_\limits{h=2}^{n-1}\left|\cot\left(\frac{a_h-a_{h-1}}{2}\right)-\cot\left(\frac{a_{h+1}-a_h}{2}\right)\right|+\\+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+|\sin a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+|\cos a_1|+1=\\=\left|\cot\left(\frac{a_2-a_1}{2}\right)-\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+\\+|\sin a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+|\cos a_1|+1=\\=\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)-\cot\left(\frac{a_2-a_1}{2}\right)\right|+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+\\+|\sin a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+|\cos a_1|+1\;,$
for all $\;n\in\mathbb{N}\setminus\{1\}.$

Property 2:
If $\;\left\{a_n\right\}_{n\in\mathbb{N}}\;$ is a sequence of real numbers such that $\;\left\{a_n-a_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonic and there exists $\;k\in\mathbb{Z}\;$ for which $\;2\pi k<a_n-a_{n-1}<2\pi+2\pi k\;\;\;\;\forall n\in\mathbb{N}\setminus\{1\}\;,\;\;\;$ then $$\left|\sum_\limits{h=1}^n \sin a_h\right|\le\frac{1}{2}\left[\; \left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)-\cot\left(\frac{a_2-a_1}{2}\right)\right|+\\+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+|\cos a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+\\+|\sin a_1|+1\;\right]$$ for all $\;n\in\mathbb{N}\setminus\{1\}.$
Proof:
By applying Prosthaphaeresis identities, we get that
$\sin a_h+\sin a_{h+1}=2\sin\left(\frac{a_{h+1}+a_h}{2}\right)\cos\left(\frac{a_{h+1}-a_h}{2}\right)=\\=2\sin\left(\frac{a_{h+1}+a_h}{2}\right)\sin\left(\frac{a_{h+1}-a_h}{2}\right)\cot\left(\frac{a_{h+1}-a_h}{2}\right)=\\=\left(\cos a_h-\cos a_{h+1}\right)\cot\left(\frac{a_{h+1}-a_h}{2}\right)\;,\;\;\text{ for all }h\in\mathbb{N}.$
Moreover,
$2\sum_\limits{h=1}^n\sin a_h=\sum_\limits{h=1}^{n-1}\left(\sin a_h +\sin a_{h+1}\right)+\sin a_1+\sin a_n=\\=\sum_\limits{h=1}^{n-1}\left(\cos a_h-\cos a_{h+1}\right)\cot\left(\frac{a_{h+1}-a_h}{2}\right)+\sin a_1+\sin a_n =\\=\sum_\limits{h=1}^{n-1}\cos a_h\cot\left(\frac{a_{h+1}-a_h}{2}\right)-\sum_\limits{h=1}^{n-1}\cos a_{h+1}\cot\left(\frac{a_{h+1}-a_h}{2}\right)+\\+\sin a_1+\sin a_n=\\=\sum_\limits{h=1}^{n-1}\cos a_h\cot\left(\frac{a_{h+1}-a_h}{2}\right)-\sum_\limits{h=2}^n\cos a_h\cot\left(\frac{a_h-a_{h-1}}{2}\right)+\\+\sin a_1+\sin a_n =\\=\sum_\limits{h=2}^{n-1}\cos a_h\left[\cot\left(\frac{a_{h+1}-a_h}{2}\right)-\cot\left(\frac{a_h-a_{h-1}}{2}\right)\right]+\\+\cos a_1\cot\left(\frac{a_2-a_1}{2}\right)-\cos a_n\cot\left(\frac{a_n-a_{n-1}}{2}\right)+\sin a_1+\sin a_n\;,\\\text{ for all }\;n\in\mathbb{N}\setminus\{1\}.$
Since the function $\;\cot\;$ is monotonic on $\;\left]\pi k,\pi+\pi k\right[\;$ and $\;\left\{a_n-a_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is a monotonic sequence such that $\;2\pi k<a_n-a_{n-1}<2\pi+2\pi k\;\;\;\;\forall n\in\mathbb{N}\setminus\{1\}\;,\;$ then the sequence $\;\left\{\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonic too.
So by taking absolute values and by noting that the sequence $\;\left\{\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonic, we get that
$2\left|\sum_\limits{h=1}^n\sin a_h\right|\le\sum_\limits{h=2}^{n-1}\left|\cot\left(\frac{a_{h+1}-a_h}{2}\right)-\cot\left(\frac{a_h-a_{h-1}}{2}\right)\right|+\\+|\cos a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+|\sin a_1|+1=\\=\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)-\cot\left(\frac{a_2-a_1}{2}\right)\right|+\left|\cot\left(\frac{a_n-a_{n-1}}{2}\right)\right|+\\+|\cos a_1|\left|\cot\left(\frac{a_2-a_1}{2}\right)\right|+|\sin a_1|+1\;,$
for all $\;n\in\mathbb{N}\setminus\{1\}.$

Corollary 1:
The sequences $\;\left\{\alpha_n=n+\sqrt{n}\right\}_{n\in\mathbb{N}}\;$ and $\left\{\beta_n=n-\sqrt{n}\right\}_{n\in\mathbb{N}}\;$ satisfy all the hypothesis of the previous properties and
$\left|\sum_\limits{h=1}^n \cos\left(h+\sqrt{h}\right)\right|<\frac{5}{2}\;,\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \sin\left(h+\sqrt{h}\right)\right|<\frac{5}{2}\;,\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \cos\left(h-\sqrt{h}\right)\right|<\frac{8}{3}\;,\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \sin\left(h-\sqrt{h}\right)\right|<\frac{23}{6}\;,\;\;\text{ for all }\;n\in\mathbb{N}.$
Proof:
$\alpha_n-\alpha_{n-1}=n+\sqrt{n}-n+1-\sqrt{n-1}=\\=1+\sqrt{n}-\sqrt{n-1}=1+\frac{1}{\sqrt{n}+\sqrt{n-1}}\;,\\\text{for all }\;n\in\mathbb{N}\setminus\{1\}.$
Hence the sequence $\;\left\{\alpha_n-\alpha_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonically decreasing and $\;0<1<\alpha_n-\alpha_{n-1}\le\sqrt{2}<\pi<2\pi\;,$
$\text{for all }\;n\in\mathbb{N}\setminus\{1\}.$
Since the function $\;\cot\;$ is monotonically decreasing on $\;\left]0,\pi\right[\;$ and $\;\left\{\alpha_n-\alpha_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is a decreasing sequence such that $\;0<\alpha_n-\alpha_{n-1}<2\pi\;\;\;\;\forall n\in\mathbb{N}\setminus\{1\}\;,\;$ then the sequence $\;\left\{\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonically increasing.
By applying the property $1$, we get that
$\left|\sum_\limits{h=1}^n \cos\alpha_h\right|\le\frac{1}{2}\left[\;\left|\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)-\cot\left(\frac{\alpha_2-\alpha_1}{2}\right)\right|+\\+\left|\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)\right|+|\sin \alpha_1|\left|\cot\left(\frac{\alpha_2-\alpha_1}{2}\right)\right|+|\cos\alpha_1|+1\;\right]=\\=\frac{1}{2}\left[\;\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)-\cot\left(\frac{\sqrt{2}}{2}\right)+\\+\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)+\sin 2\cot\left(\frac{\sqrt{2}}{2}\right)-\cos 2+1\;\right]=\\=\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)+\frac{1}{2}\left(\sin 2-1\right)\cot\left(\frac{\sqrt{2}}{2}\right)+\frac{1}{2}\left(1-\cos 2\right)<\\<\cot\left(\frac{1}{2}\right)+\frac{1}{2}\left(\sin 2-1\right)\cot\left(\frac{\sqrt{2}}{2}\right)+\frac{1}{2}\left(1-\cos 2\right)<\frac{5}{2}\;,$
for all $\;n\in\mathbb{N}\setminus\{1\}.$
Therefore,
$\left|\sum_\limits{h=1}^n \cos\left(h+\sqrt{h}\right)\right|<\frac{5}{2}\;,\;\;\text{ for all }\;n\in\mathbb{N}.$
And by applying the property $2$, we get that
$\left|\sum_\limits{h=1}^n \sin\alpha_h\right|\le\frac{1}{2}\left[\;\left|\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)-\cot\left(\frac{\alpha_2-\alpha_1}{2}\right)\right|+\\+\left|\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)\right|+|\cos \alpha_1|\left|\cot\left(\frac{\alpha_2-\alpha_1}{2}\right)\right|+|\sin\alpha_1|+1\;\right]=\\=\frac{1}{2}\left[\;\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)-\cot\left(\frac{\sqrt{2}}{2}\right)+\\+\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)-\cos 2\cot\left(\frac{\sqrt{2}}{2}\right)+\sin 2+1\;\right]=\\=\cot\left(\frac{\alpha_n-\alpha_{n-1}}{2}\right)-\frac{1}{2}\left(1+\cos 2\right)\cot\left(\frac{\sqrt{2}}{2}\right)+\frac{1}{2}\left(1+\sin 2\right)<\\<\cot\left(\frac{1}{2}\right)-\frac{1}{2}\left(1+\cos 2\right)\cot\left(\frac{\sqrt{2}}{2}\right)+\frac{1}{2}\left(1+\sin 2\right)<\frac{5}{2}\;,$
for all $\;n\in\mathbb{N}\setminus\{1\}.$
Therefore,
$\left|\sum_\limits{h=1}^n \sin\left(h+\sqrt{h}\right)\right|<\frac{5}{2}\;,\;\;\text{ for all }\;n\in\mathbb{N}.$
Moreover,
$\beta_n-\beta_{n-1}=n-\sqrt{n}-n+1+\sqrt{n-1}=\\=1-\sqrt{n}+\sqrt{n-1}=1-\frac{1}{\sqrt{n}+\sqrt{n-1}}\;,\\\text{for all }\;n\in\mathbb{N}\setminus\{1\}.$
Hence the sequence $\;\left\{\beta_n-\beta_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonically increasing and $\;0<2-\sqrt{2}\le\beta_n-\beta_{n-1}<1<\pi<2\pi\;,$
$\text{for all }\;n\in\mathbb{N}\setminus\{1\}.$
Since the function $\;\cot\;$ is monotonically decreasing on $\;\left]0,\pi\right[\;$ and $\;\left\{\beta_n-\beta_{n-1}\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is an increasing sequence such that $\;0<\beta_n-\beta_{n-1}<2\pi\;\;\;\;\forall n\in\mathbb{N}\setminus\{1\}\;,\;$ then the sequence $\;\left\{\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)\right\}_{n\in\mathbb{N}\setminus\{1\}}\;$ is monotonically decreasing.
By applying the property $1$, we get that
$\left|\sum_\limits{h=1}^n \cos\beta_h\right|\le\frac{1}{2}\left[\;\left|\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)-\cot\left(\frac{\beta_2-\beta_1}{2}\right)\right|+\\+\left|\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)\right|+|\sin \beta_1|\left|\cot\left(\frac{\beta_2-\beta_1}{2}\right)\right|+|\cos\beta_1|+1\;\right]=\\=\frac{1}{2}\left[\;\cot\left(\frac{2-\sqrt{2}}{2}\right)-\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)+\\+\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)+\sin 0\cot\left(\frac{2-\sqrt{2}}{2}\right)+\cos0+1\;\right]=\\=1+\frac{1}{2}\cot\left(\frac{2-\sqrt{2}}{2}\right)<\frac{8}{3}\;,$
for all $\;n\in\mathbb{N}\setminus\{1\}.$
Therefore,
$\left|\sum_\limits{h=1}^n \cos\left(h-\sqrt{h}\right)\right|<\frac{8}{3}\;,\;\;\text{ for all }\;n\in\mathbb{N}.$
And by applying the property $2$, we get that
$\left|\sum_\limits{h=1}^n \sin\beta_h\right|\le\frac{1}{2}\left[\;\left|\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)-\cot\left(\frac{\beta_2-\beta_1}{2}\right)\right|+\\+\left|\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)\right|+|\cos\beta_1|\left|\cot\left(\frac{\beta_2-\beta_1}{2}\right)\right|+|\sin\beta_1|+1\;\right]=\\=\frac{1}{2}\left[\;\cot\left(\frac{2-\sqrt{2}}{2}\right)-\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)+\\+\cot\left(\frac{\beta_n-\beta_{n-1}}{2}\right)+\cos 0\cot\left(\frac{2-\sqrt{2}}{2}\right)+\sin 0+1\;\right]=\\=\frac{1}{2}+\cot\left(\frac{2-\sqrt{2}}{2}\right)<\frac{23}{6}\;,$
for all $\;n\in\mathbb{N}\setminus\{1\}.$
Therefore,
$\left|\sum_\limits{h=1}^n \sin\left(h-\sqrt{h}\right)\right|<\frac{23}{6}\;,\;\;\text{ for all }\;n\in\mathbb{N}.$

Corollary 2:
$\left|\sum_\limits{h=1}^n \sin h\sin\sqrt{h}\right|<\frac{31}{12}<\frac{13}{5}\;,\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \cos h\cos\sqrt{h}\right|<\frac{31}{12}<\frac{13}{5}\;,\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \sin h\cos\sqrt{h}\right|<\frac{19}{6}<\frac{16}{5}\;,\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \cos h\sin\sqrt{h}\right|<\frac{19}{6}<\frac{16}{5}\;,\;\;\text{ for all }\;n\in\mathbb{N}.$
Proof:
Using the results of Corollary 1, we get that
$\left|\sum_\limits{h=1}^n \sin h\sin\sqrt{h}\right|=\frac{1}{2}\left|\sum_\limits{h=1}^n \left[\cos\left(h-\sqrt{h}\right)-\cos\left(h+\sqrt{h}\right)\right]\right|=\\=\frac{1}{2}\left|\sum_\limits{h=1}^n\cos\left(h-\sqrt{h}\right)-\sum_\limits{h=1}^n\cos\left(h+\sqrt{h}\right)\right|\le\\\le\frac{1}{2}\left[\;\left|\sum_\limits{h=1}^n\cos\left(h-\sqrt{h}\right)\right|+\left|\sum_\limits{h=1}^n\cos\left(h+\sqrt{h}\right)\right|\;\right]<\\<\frac{1}{2}\left[\frac{8}{3}+\frac{5}{2}\right]=\frac{31}{12}<\frac{13}{5}\;,\;\;\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \cos h\cos\sqrt{h}\right|=\frac{1}{2}\left|\sum_\limits{h=1}^n \left[\cos\left(h+\sqrt{h}\right)+\cos\left(h-\sqrt{h}\right)\right]\right|=\\=\frac{1}{2}\left|\sum_\limits{h=1}^n\cos\left(h+\sqrt{h}\right)+\sum_\limits{h=1}^n\cos\left(h-\sqrt{h}\right)\right|\le\\\le\frac{1}{2}\left[\;\left|\sum_\limits{h=1}^n\cos\left(h+\sqrt{h}\right)\right|+\left|\sum_\limits{h=1}^n\cos\left(h-\sqrt{h}\right)\right|\;\right]<\\<\frac{1}{2}\left[\frac{5}{2}+\frac{8}{3}\right]=\frac{31}{12}<\frac{13}{5}\;,\;\;\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \sin h\cos\sqrt{h}\right|=\frac{1}{2}\left|\sum_\limits{h=1}^n \left[\sin\left(h+\sqrt{h}\right)+\sin\left(h-\sqrt{h}\right)\right]\right|=\\=\frac{1}{2}\left|\sum_\limits{h=1}^n\sin\left(h+\sqrt{h}\right)+\sum_\limits{h=1}^n\sin\left(h-\sqrt{h}\right)\right|\le\\\le\frac{1}{2}\left[\;\left|\sum_\limits{h=1}^n\sin\left(h+\sqrt{h}\right)\right|+\left|\sum_\limits{h=1}^n\sin\left(h-\sqrt{h}\right)\right|\;\right]<\\<\frac{1}{2}\left[\frac{5}{2}+\frac{23}{6}\right]=\frac{19}{6}<\frac{16}{5}\;,\;\;\;\;\text{ for all }\;n\in\mathbb{N}\;,$
$\left|\sum_\limits{h=1}^n \cos h\sin\sqrt{h}\right|=\frac{1}{2}\left|\sum_\limits{h=1}^n \left[\sin\left(h+\sqrt{h}\right)-\sin\left(h-\sqrt{h}\right)\right]\right|=\\=\frac{1}{2}\left|\sum_\limits{h=1}^n\sin\left(h+\sqrt{h}\right)-\sum_\limits{h=1}^n\sin\left(h-\sqrt{h}\right)\right|\le\\\le\frac{1}{2}\left[\;\left|\sum_\limits{h=1}^n\sin\left(h+\sqrt{h}\right)\right|+\left|\sum_\limits{h=1}^n\sin\left(h-\sqrt{h}\right)\right|\;\right]<\\<\frac{1}{2}\left[\frac{5}{2}+\frac{23}{6}\right]=\frac{19}{6}<\frac{16}{5}\;,\;\;\;\;\text{ for all }\;n\in\mathbb{N}\;.$

Corollary 3:
$\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\ldots+\sin n\sin\sqrt{n}}{n^\gamma}=0$
for any $\;\gamma>0.$
Proof:
Since $$-\frac{31}{12 n^\gamma}<\frac{\sum_\limits{h=1}^n \sin h\sin\sqrt{h}}{n^\gamma}<\frac{31}{12 n^\gamma}\;\;\;\;\text{ for all }\;n\in\mathbb{N}$$ and $\;\lim_\limits{n\to\infty}\left(-\frac{31}{12 n^\gamma}\right)=0\;,\;\;\lim_\limits{n\to\infty}\frac{31}{12 n^\gamma}=0\;,$
by applying the squeeze theorem, we get that
$\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\ldots+\sin n\sin\sqrt{n}}{n^\gamma}=0\;.$