We will use the following (well known?) fact:

*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 might be going back as far as to Dirichlet, though I wasn't able to find an exact reference. Also it is possible that Ramanujan new it, because he studied sums of this kind (see periodic zeta functions in his Lost Notebook).

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 $f\equiv 1$.