Solve that $\displaystyle\lim_{n\to\infty}\frac{\sum_{k=1}^n\sin\sqrt k}{n}=?$ I guess$\displaystyle\lim_{n\to\infty}\frac{\sum_{k=1}^n\sin\sqrt k}{n}=0$. 
If it converges to zero, it is very slow.
Even I make $n=1,000,000$, the result is $-0.00112289$.
Will anyone help me out?
Thanks very much!
 A: By Abel's summation we have $$\sum_{k\leq n}\sin\left(\sqrt{k}\right)=\sum_{k\leq n}1\cdot\sin\left(\sqrt{k}\right)=n\sin\left(\sqrt{n}\right)-\int_{1}^{n}\frac{\left\lfloor t\right\rfloor \cos\left(\sqrt{t}\right)}{2\sqrt{t}}dt$$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. Now using $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we get $$=n\sin\left(\sqrt{n}\right)-\frac{1}{2}\int_{1}^{n}\sqrt{t}\cos\left(\sqrt{t}\right)dt+O\left(\sin\left(\sqrt{n}\right)\right)
 $$ and note that the integral is quite easy to calculate. Take $\sqrt{t}=u
 $ and integrate by parts to get $$\frac{1}{2}\int_{1}^{n}\sqrt{t}\cos\left(\sqrt{t}\right)dt=\int_{1}^{\sqrt{n}}u^{2}\cos\left(u\right)du=n\sin\left(\sqrt{n}\right)-2\int_{1}^{\sqrt{n}}u\sin\left(u\right)du+O\left(1\right)=
 $$ $$ =n\sin\left(\sqrt{n}\right)+2\sqrt{n}\cos\left(\sqrt{n}\right)-2\int_{1}^{\sqrt{n}}\cos\left(u\right)du+O\left(1\right)=
 $$ $$=n\sin\left(\sqrt{n}\right)+2\sqrt{n}\cos\left(\sqrt{n}\right)-2\sin\left(\sqrt{n}\right)+O\left(1\right)
 $$ hence $$\sum_{k\leq n}\sin\left(\sqrt{k}\right)=-2\sqrt{n}\cos\left(\sqrt{n}\right)+O\left(\sin\left(\sqrt{n}\right)\right)=-2\sqrt{n}\cos\left(\sqrt{n}\right)+O\left(1\right)
 $$ and so the limit is $0$ and this also confirm the result of Sangchul Lee.
A: I have another idea.
Let$$a_k := \int_{k}^{k+1} \sin (\sqrt{x}) dx - \sin (\sqrt{k}). $$
and $f(x) :=\sin (\sqrt{x})$. Since $f'(x)=\frac{\cos (\sqrt{x})}{2\sqrt{x}}$.
By the mean value theorem, for some $x_k \in [k,k+1]$,  $$\int_{k}^{k+1} \sin (\sqrt{x}) dx =g(x_k).$$ Then, $|g(x_k)-g(k)| \leq \sup \limits_{t \in [k,k+1]} |g'(t)|$. Thus, there is some $y_k \in [k,k+1)$ s.t.
$a_k=f'(y_k)$, and $$|a_k| \leq \frac{1}{2\sqrt{k}}.$$
As a result, $$|\sum_{k=1}^n\sin\sqrt k - \int_{1}^{n} \sin (\sqrt{x}) dx| \\
\leq |\sum_{k=1}^n \frac{1}{2\sqrt k}|< \sqrt n$$
Now, consider that
$$\int_{1}^{n} \sin (\sqrt{x}) dx
=2(\sin (\sqrt x)- \sqrt x \cos (\sqrt x))|_{1}^{n} \\
=2(\sin (\sqrt n)- \sqrt x \cos (\sqrt n) + C)$$
Where $C=\cos 1-\sin 1$.
While
$$|\sin (\sqrt n)- \sqrt x \cos (\sqrt n)| \\
=\sqrt n|\cos (\sqrt n)- \frac{\cos (\sqrt n)}{\sqrt n}| \\
<\sqrt n(|\cos (\sqrt n)|+ |\frac{\cos (\sqrt n)}{\sqrt n}|) \\
<\sqrt n(1+ \frac{1}{\sqrt n})
=\sqrt n +1.$$
As a result,
$$\sum_{k=1}^n \sin\sqrt k < \sqrt n+ |\int_{1}^{n} \sin (\sqrt{x}) dx| \\
< 3\sqrt n+ 2+2C$$
Since $\lim \limits_{n \to \infty} \frac{\sqrt n}{n}=0$, by squeeze theorem, the limit required is $0$.
