I have encountered a limiting process on some infinite series. So, I would like to ask:
QUESTION. Assume $n$ is an even positive integer. Is this true? $$\lim_{r\rightarrow1^{-}}\sum_{j=1}^{\infty}\left(\sqrt{\frac2n}-\cos\left(\frac{\pi j^2}n\right) -\sin\left(\frac{\pi j^2}n\right)\right)r^j=\frac12\left(1-\sqrt{\frac2n}\right).$$
We will use the following well known fact (e.g., see sections 1.1 and 1.2 in this article):
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 $f\equiv 1$.
