# 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? – Angelo Aug 12 '20 at 16:23
• @NamburuKarthik we have to prove it, not speculate an answer – Anindya Prithvi Aug 12 '20 at 16:24
• And i think you'd like to use Cauchy-Schwartz inequality in your last two equations, it would be sufficient – Anindya Prithvi Aug 12 '20 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. – Angelo Aug 12 '20 at 16:51
• Look at this i.stack.imgur.com/zn0eA.png. So “desmos” give us a confirmation that $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\ldots+\sin n\sin\sqrt{n}}{n}=0$ – Angelo Aug 12 '20 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 $o(1) ~ O(1/n)$? – MoonKnight Aug 12 '20 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 $o(1)$? – Angelo Aug 12 '20 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. – Mark Viola Aug 12 '20 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 – Conrad Aug 12 '20 at 20:25
• Thanks Conrad! And I hope that you and your family are staying safe and healthy. – Mark Viola Aug 12 '20 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? – Angelo Aug 12 '20 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 $e^{i(k+\sqrt{k})}=e^{-2\pi i g(k)}$. – Angelo Aug 12 '20 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. – Mark Viola Aug 12 '20 at 20:11
• You are right - I switched the estimates, will correct – Conrad Aug 12 '20 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 $$S_2(n)=\sum_{k=1}^{n}e^{2\pi i g(k)}=\sum_1^{n}1/2(1+i\cot \pi q(k))(e^{2\pi i g(k)}-e^{2\pi i g(k+1)})=$$ $$=i/2\sum_{k=2}^{n} e^{2\pi i g(k)}(\cot \pi q(k)-\cot \pi q(k-1))+1/2(1+i\cot \pi q(1))e^{2\pi i g(1)}-1/2(1+i\cot \pi q(n))e^{2\pi i g(n+1)}$$ – Angelo Aug 13 '20 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. – Angelo Aug 14 '20 at 18:27
• Oops ... I missed the factor of $\frac12$ in the argument. Please accept my apology. – Mark Viola Aug 14 '20 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. – Angelo Aug 14 '20 at 18:34
• You are welcome. If you notice some mistake, please tell me it. – Angelo Aug 14 '20 at 18:37
• This looks good Angelo. With $a_n=n\pm \sqrt{n}$, we have for $n\ne1$ $$\frac12\left(a_n-a_{n-1}\right)=\frac12\left(1\pm\frac1{\sqrt{n}+\sqrt{n-1}}\right)$$ – Mark Viola Aug 14 '20 at 18:44