We will use the following well known fact (e.g., see sections 1.1 and 1.2 in this [article][1]):

*Given $f(x)$ with period $1$, its Fourier series
$$
f(x)=\sum_{j=0}^\infty a_j\cos(2\pi jx),
$$
and a positive integer $n\in\mathbb{N}$, one has*
$$
\sum_{j=0}^\infty a_je^{\pi i j^2/n}=\frac{e^{\pi i (1-n)/4}}{\sqrt{n}}\sum _{j=1}^n(-1)^jf\Big(\frac{1}{2}-\frac{j}{n}\Big)e^{-\pi i j^2/n}.
$$

The idea behind this fact is due to Dirichlet (see Andrews, Askey and Roy's book).

Now take $a_j=r^j$, $|r|<1$. Then
$$
f(x)=\frac{1-r\cos(2\pi x)}{1-2r\cos(2\pi x)+r^2}.
$$

This allows one to calculate the part of OP's series that contain cosine and sine terms. The part that contains $\sqrt{2/n}$ is trivial.

Now that the series is reduced to a finite sum, one can put $r\to 1-0$.


In the resulting finite sum, the singular terms $(1-r)^{-1}$ come from the terms in the series that contain $\sqrt{2/n}$, and one is contained in the finite sum from $f(0)$ (the term $j=n$), and they would cancel each other.

Since for $x\neq 0$
$$
\lim_{r\to 1}f(x)=1/2,\quad x\neq 0, 
$$
OP's claim reduces to a calculation of a Gauss sum. However, there is no need to calculate this Gauss sum explicitly, because its value follows from the general formula if one takes $f\equiv 1$.


  [1]: https://arxiv.org/abs/2404.10805