How to show that $\sum_{0}^{M} \sin(\sqrt{n^{2}+1}x) $ is bounded $\forall x \in (0,\pi)$ for any constant $M$ I tried to estimate it like $Const\cdot \sum_{0}^{M} \sin(nx)$ ($M$ is arbitrary). And the last one is obviously bounded. But didn't find any constant. Any hints ?
 A: The supremum of that sum of functions in the interval $x\in(0,\pi)$ grows linearly with $M$, so a bound independent of $M$ cannot exist.
The problem appears at the low end of that interval. Let $x=\pi/(2M)$. Then $\sqrt{n^2+1}x$ always belongs to a suitable interval like $I=[0,9/5]$ (we may need to exclude a few small values of $M$). In the interval $I$ we have the lower bound $\sin x>x/2$. Therefore
$$
\begin{aligned}
\sum_{n=0}^M\sin(\sqrt{n^2+1}x)&\ge \sum_{n=0}^M\frac{\sqrt{n^2+1}x}2\\
&\ge\sum_{n=0}^M\frac{nx}2\\
&=\frac{M(M+1)x}4=\frac{\pi (M+1)}8.
\end{aligned}
$$
This kills all hope for a bound that holds for all $M$ and all $x\in(0,\pi)$.
On the other hand plotting a few such sums suggests that the problem is only at the low end. Here is what the plot looks like with $M=50$,

and here's how it changes when $M=100$.

A surprisingly similar overall shape if you ask me! I guess the fact that the periods of the summands are linearly independent forces significant cancellation to take place when $x$ is large enough! After viewing a few more such plots it would not necessarily surprise me if to each interval $x\in(\epsilon,\pi)$, $\epsilon>0$, there were a bound that would not depend on $M$ (but by the above example it would necessarily depend on $\epsilon$).
A: Since $-1 \leq \sin(y) \leq 1$ for every $y,$ we have that $-M \leq \displaystyle \sum_{n=0}^M \sin(\sqrt{n^2+1}x) \leq M.$ 
