Bound on $\sum\limits_{n=0}^{x}{\sin{\sqrt{n}}}$ Using Desmos and Mathematica, I was able to find a function $g(x)$ that seemingly estimated the function 
$$f(x)=\sum_{n=0}^{x}{\sin{\sqrt{n}}}$$ 
I found that 
$${g(x)=2\sqrt{x}*\sin{\left({{\sqrt{4x+{\pi}^2}-\pi}\over{2}}\right)}}\approx f(x)$$
Furthermore, I found that the difference between these two functions, or $|{f(x)-g(x)}|$, never seemed to exceed some constant $C\approx 0.464568$ 
Is there some way to prove the conjecture below?
$g(x)<f(x)<(g(x)+C)$ for all $x>14$
Or perhaps a weaker version considering I have no good definition for $C$:
$g(x)<f(x)<(g(x)+{1\over2})$ for all $x>14$.
Further questions:
Is there a closed form for $C$?
Is there a closed form for $f(x)$?

(Too long for a comment)

We have
\begin{align*}
f(x)
= \sum_{k=1}^{n} \int_{0}^{\sqrt{k}} \cos x \, dx
= \int_{0}^{\infty} \left( \sum_{k=1}^{n} \mathbf{1}_{\{ x < \sqrt{k}\}} \right) \cos x \, dx.
\end{align*}
Notice that $\sum_{k=1}^{n} \mathbf{1}_{\{ x < \sqrt{k}\}} = (n - \lfloor x^2 \rfloor)_+$, where $a_+ = \max\{a, 0\}$ denotes the positive part of $a$. So
\begin{align*}
f(x)
&= \int_{0}^{\sqrt{n}} (n - \lfloor x^2 \rfloor) \cos x \, dx \\
&= \int_{0}^{\sqrt{n}} \left(n - x^2 + \frac{1}{2} \right) \cos x \, dx + \int_{0}^{\sqrt{n}} \left(x^2 - \lfloor x^2 \rfloor - \frac{1}{2} \right) \cos x \, dx
\end{align*}
The first integral is easy to evaluate:
$$ \int_{0}^{\sqrt{n}} \left(n - x^2 + \frac{1}{2} \right) \cos x \, dx
= -2\sqrt{n}\cos\sqrt{n} + \frac{5}{2}\sin\sqrt{n}. $$
To evaluate the second integral, we substitute $x^2 \mapsto x$ to write
\begin{align*}
\int_{0}^{\sqrt{n}} \left(x^2 - \lfloor x^2 \rfloor - \frac{1}{2} \right) \cos x \, dx
&= \sum_{k=0}^{n-1} \int_{0}^{1} \left(x - \frac{1}{2}\right) \frac{\cos\sqrt{x+k}}{2\sqrt{x+k}} \, dx \\
&= \sum_{k=0}^{n-1} \int_{0}^{\frac{1}{2}} x \left( \frac{\cos\sqrt{k+\frac{1}{2}+x}}{2\sqrt{k+\frac{1}{2}+x}} - \frac{\cos\sqrt{k+\frac{1}{2}-x}}{2\sqrt{k+\frac{1}{2}-x}} \right) \, dx
\end{align*}
Although I cannot prove at this point, numerical evidence suggests that this integral converge as $n\to\infty$. Assuming this, it follows that

Conjecture. $c = \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left(x^2 - \lfloor x^2 \rfloor - \frac{1}{2} \right) \cos x \, dx $ converges. Consequently,
$$ f(n) = -2\sqrt{n}\cos\sqrt{n} + \frac{5}{2}\sin\sqrt{n} + c + o(1) $$

Indeed, the following figure is the plot of the difference $f(n) - \left( -2\sqrt{n}\cos\sqrt{n} + \frac{5}{2}\sin\sqrt{n} \right)$ for $n \leq 10^4$.

