How to prove that a sum: $\sum_{n=1}^\infty \cos(n\sqrt 2)$ is uniformly bounded? How to prove that a sum:
$$\sum_{n=1}^\infty \cos(n\sqrt 2)$$
is uniformly bounded?
That problem is connected to an answer to other question that I have asked before:
Convergence of $\sum_{n=0}^{\infty}\frac{cos(n \sqrt{2})}{\sqrt{n}}$. Is my thinking correct?

I'll provide two methods, one using complex numbers and one using sum-of-angle formulas. Let $b=\sqrt 2$.

Note that by Euler's formula:
\begin{align*}
\sum_{n=1}^N\cos\left(bn\right)&=\Re\sum_{n=1}^N\left(\cos\left(bn\right)+i\sin\left(bn\right)\right)\\
&=\Re\sum_{n=1}^Ne^{ibn}\\
&=\Re\frac{e^{ib(N+1)}-e^{ib}}{e^{ib}-1}
\end{align*}
Here $e^{ib}-1\ne 0$ and $e^{ib(N+1)}$ is uniformly bounded, so the partial sum is also uniformly bounded.

To use trigonometry, you want to recall formulas for $f\left((n\pm 1)b\right)$ where $f$ is $\sin$ or $\cos$. In this case $\sin$ works:
\begin{align*}
\Delta_n&=\sin\left((n+1)b\right)-\sin\left((n-1)b\right)\\
&=\sin(nb)\cos(b)+\cos(nb)\sin(b)-\left\{\sin(nb)\cos(b)-\cos(nb)\sin(b)\right\}\\
&=2\cos(nb)\sin(b)
\end{align*}
Therefore we have a telescoping sum:
\begin{align*}
\sum_{n=1}^N\cos\left(bn\right)&=\frac{1}{2\sin\left(b\right)}\sum_{n=1}^N\left(\sin\left((n+1)b\right)-\sin\left((n-1)\right)b\right)\\
&=\frac{\sin\left((N+1)b\right)+\sin\left(Nb\right)-\sin\left(b\right)-\sin\left(0\right)}{2\sin(b)}
\end{align*}
Now again $\sin(b)\ne 0$ and $\sin((N+1)b)+\sin(Nb)$ is uniformly bounded, so the partial sum is uniformly bounded.
