# 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 $$\frac{1}{2}$$.

• Yes, I know it but how can I prove that the limit is zero formally and rigorously? Aug 12, 2020 at 16:23
• @NamburuKarthik we have to prove it, not speculate an answer Aug 12, 2020 at 16:24
• And i think you'd like to use Cauchy-Schwartz inequality in your last two equations, it would be sufficient Aug 12, 2020 at 16:28
• Because you have to apply Cauchy-Schwartz inequality to the numerators otherwise you would get $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\ldots+\sin n\sin\sqrt{n}}{n^2}$ which is not the limit I wish to. Aug 12, 2020 at 16:51
• Look at this i.stack.imgur.com/zn0eA.png. So “desmos” give us a confirmation that $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\ldots+\sin n\sin\sqrt{n}}{n}=0$ Aug 12, 2020 at 18:18

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 $$\limsup_{n\to \infty}\frac{S_n}{\sqrt n}\le 2\csc(1/2)$$

• Sorry, could you elaborate why the difference between $S_n$ and the integral is $o(1)$? btw, I assume $o(1) ~ O(1/n)$? Aug 12, 2020 at 16:54
• Could you explain me why $\frac{1}{n}\int_0^n\left(x-\lfloor x\rfloor\right)\left(\cos x\sin\sqrt{x}+\frac{\sin x\cos\sqrt{x}}{2\sqrt{x}}\right)dx$ is $o(1)$? Aug 12, 2020 at 17:33
• @Angelo Rather than elaborate on the former solution, I've taken another approach. This new approach relies only on summation by parts and some simple estimates. Let me know your thoughts. Aug 12, 2020 at 20:12
• Nice solution too - uses the estimate on the square root only which indeed is simpler and that is enough here (basically the $\sin n$ coefficients can be any $a_n$ with uniformly bounded sum and your solution works with a square root bound while use that $x$ dominated $\sqrt x$ essentially but get a better bound Aug 12, 2020 at 20:25
• Thanks Conrad! And I hope that you and your family are staying safe and healthy. Aug 12, 2020 at 20:26

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 $$0k$$ 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!

• Why have you written $1/(4\pi) \le g'(x) \le 1/(2\pi)$ ? I think that $1/(2\pi) \le g'(x) \le 3/(4\pi)$, am I wrong? Aug 12, 2020 at 19:49
• $g’(x)$ is decreasing and if we replace $g(x)$ by $-g(x)$ then $-\frac{3}{4\pi}\le q(n)\le -\frac{1}{2\pi}$ and $e^{i(k+\sqrt{k})}=e^{-2\pi i g(k)}$. Aug 12, 2020 at 20:09
• @conrad Nice solution. I've posted another one that relies only on summation by parts and some simple estimates. Let me know your thoughts. Aug 12, 2020 at 20:11
• You are right - I switched the estimates, will correct Aug 12, 2020 at 20:12
• @Conrad, you wrote that $$S_2(n)=\sum_{k=1}^{n}e^{2\pi i g(k)}=\sum_1^{n}1/2(1+i\cot \pi q(k))(e^{2\pi i g(k)}-e^{2\pi i g(k+1)})=$$ $$=i/2\sum_{k=2}^{n-1}e^{2\pi i g(k)}(\cot \pi q(k)-\cot \pi q(k+1))+1/2(1+i\cot \pi q(1))e^{2\pi i g(1)}-(1/2)(1+i\cot \pi q(n))e^{2\pi i g(n+1)}$$ but I have obtained that $$S_2(n)=\sum_{k=1}^{n}e^{2\pi i g(k)}=\sum_1^{n}1/2(1+i\cot \pi q(k))(e^{2\pi i g(k)}-e^{2\pi i g(k+1)})=$$ $$=i/2\sum_{k=2}^{n} e^{2\pi i g(k)}(\cot \pi q(k)-\cot \pi q(k-1))+1/2(1+i\cot \pi q(1))e^{2\pi i g(1)}-1/2(1+i\cot \pi q(n))e^{2\pi i g(n+1)}$$ Aug 13, 2020 at 10:17

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 $$\;2\pi k then $$\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 $$\;2\pi k then the sequence $$\;\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 $$\;2\pi k then $$\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 $$\;2\pi k then the sequence $$\;\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

$$\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\ldots+\sin n\sin\sqrt{n}}{n^\gamma}=0\;.$$

• If $a_n-a_{n-1}$ is an odd multiple of $\pi$, then $\cot\left(\frac{a_n-a_{n-1}}{2}\right)$ is zero and it improves the estimate. Anyway my approach is the same of Conrad’s but I have only taken into account the real part of the complex numbers. Aug 14, 2020 at 18:27
• Oops ... I missed the factor of $\frac12$ in the argument. Please accept my apology. Aug 14, 2020 at 18:32
• Since $\;2\pi k<a_n-a_{n-1}<2\pi+2\pi k$, then $\;\pi k<\frac{a_n-a_{n-1}}{2}<\pi+\pi k$, hence the function cotangent is defined for all $n\ge2$. Look at the hypothesis. Aug 14, 2020 at 18:34
• You are welcome. If you notice some mistake, please tell me it. Aug 14, 2020 at 18:37
• This looks good Angelo. With $a_n=n\pm \sqrt{n}$, we have for $n\ne1$ $$\frac12\left(a_n-a_{n-1}\right)=\frac12\left(1\pm\frac1{\sqrt{n}+\sqrt{n-1}}\right)$$ Aug 14, 2020 at 18:44