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We will use the following well known fact (two proofs of this fact can be found at the end of the post):

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 $f\equiv 1$.

EDIT: 1st proof of the general fact using multisection. $$ \sum_{j\in\mathbb{Z}} a_je^{\pi i j^2/n}=\sum_{s=1}^ne^{\pi i s^2/n}\sum_{k\in\mathbb{Z}} a_{s+kn}(-1)^k\\ =\sum_{s=1}^ne^{\pi i s^2/n}\frac{1}{n}\sum_{j=1}^ne^{2\pi isj/n}f(1/2-j/n)\\ =\frac{1}{n}\sum_{j=1}^nf(1/2-j/n)\sum_{s=1}^ne^{2\pi isj/n}e^{\pi i s^2/n}\\ =\frac{1}{n}\sum_{j=1}^nf(1/2-j/n)e^{-\pi i j^2/n}\sum_{s=1}^ne^{\pi i (s+j)^2/n} $$

The inner sum here is a Gauss sum and obviously it does not depend on $j$. Its value is known from other sources. The proof below avoids calculation of Gauss sums, but probably will be called unsatisfiable by some people.

2nd proof of the general fact using Poisson summation formula (see transition from the 3rd line to the 4th, where it is effectively applied to a function which is $0$ outside $[-1/2,1/2]$).

\begin{align} \frac{e^{\pi i/4}}{\sqrt{a}}\sum_{j=-\infty}^\infty a_je^{-\pi i j^2/a}&=\int_{-\infty}^\infty f(x) e^{\pi i a x^2}dx\\ &=\int_{-1/2}^{1/2} f(x) \sum_{n=-\infty}^\infty e^{\pi i a (x+n)^2}dx\\ &=\sum_{n=-\infty}^\infty e^{\pi i a n}\int_{-1/2}^{1/2} f(x)e^{\pi i a x^2} e^{2\pi i a n x}dx\\ &=\frac{1}{a}\sum _{n=0}^a f\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i a\left(\frac{1}{2}-\frac{n}{a}\right)^2}\\ &=\frac{e^{\pi i a/4}}{a}\sum _{n=0}^a (-1)^nf\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i n^2/a} \end{align}

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 $f\equiv 1$.

We will use the following well known fact (two proofs of this fact can be found at the end of the post):

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 $f\equiv 1$.

EDIT: 1st proof of the general fact using multisection. $$ \sum_{j\in\mathbb{Z}} a_je^{\pi i j^2/n}=\sum_{s=1}^ne^{\pi i s^2/n}\sum_{k\in\mathbb{Z}} a_{s+kn}e^{\pi i j^2/n}\\ =\sum_{s=1}^ne^{\pi i s^2/n}\frac{1}{n}\sum_{j=1}^ne^{2\pi isj/n}f(1/2-j/n)\\ =\frac{1}{n}\sum_{j=1}^nf(1/2-j/n)\sum_{s=1}^ne^{2\pi isj/n}e^{\pi i s^2/n} $$

The inner sum here is a Gauss sum and its value is known from other sources. The proof below avoids calculation of Gauss sums, but probably will be called unsatisfiable by some.

2nd proof of the general fact using Poisson summation formula (see transition from the 3rd line to the 4th).

\begin{align} \frac{1}{\sqrt{a}}\sum_{j=-\infty}^\infty a_je^{-\pi i j^2/a}&=\int_{-\infty}^\infty f(x) e^{\pi i a x^2}dx\\ &=\int_{-1/2}^{1/2} f(x) \sum_{n=-\infty}^\infty e^{\pi i a (x+n)^2}dx\\ &=\sum_{n=-\infty}^\infty e^{\pi i a n}\int_{-1/2}^{1/2} f(x)e^{\pi i a x^2} e^{2\pi i a n x}dx\\ &=\frac{1}{a}\sum _{n=0}^a f\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i a\left(\frac{1}{2}-\frac{n}{a}\right)^2}\\ &=\frac{e^{\pi i (a-1)/4}}{a}\sum _{n=0}^a (-1)^nf\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i n^2/a} \end{align}

We will use the following well known fact (two proofs of this fact can be found at the end of the post):

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 $f\equiv 1$.

EDIT: 1st proof of the general fact using multisection. $$ \sum_{j\in\mathbb{Z}} a_je^{\pi i j^2/n}=\sum_{s=1}^ne^{\pi i s^2/n}\sum_{k\in\mathbb{Z}} a_{s+kn}(-1)^k\\ =\sum_{s=1}^ne^{\pi i s^2/n}\frac{1}{n}\sum_{j=1}^ne^{2\pi isj/n}f(1/2-j/n)\\ =\frac{1}{n}\sum_{j=1}^nf(1/2-j/n)\sum_{s=1}^ne^{2\pi isj/n}e^{\pi i s^2/n}\\ =\frac{1}{n}\sum_{j=1}^nf(1/2-j/n)e^{-\pi i j^2/n}\sum_{s=1}^ne^{\pi i (s+j)^2/n} $$

The inner sum here is a Gauss sum and obviously it does not depend on $j$. Its value is known from other sources. The proof below avoids calculation of Gauss sums, but probably will be called unsatisfiable by some people.

2nd proof of the general fact using Poisson summation formula (see transition from the 3rd line to the 4th, where it is effectively applied to a function which is $0$ outside $[-1/2,1/2]$).

\begin{align} \frac{e^{\pi i/4}}{\sqrt{a}}\sum_{j=-\infty}^\infty a_je^{-\pi i j^2/a}&=\int_{-\infty}^\infty f(x) e^{\pi i a x^2}dx\\ &=\int_{-1/2}^{1/2} f(x) \sum_{n=-\infty}^\infty e^{\pi i a (x+n)^2}dx\\ &=\sum_{n=-\infty}^\infty e^{\pi i a n}\int_{-1/2}^{1/2} f(x)e^{\pi i a x^2} e^{2\pi i a n x}dx\\ &=\frac{1}{a}\sum _{n=0}^a f\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i a\left(\frac{1}{2}-\frac{n}{a}\right)^2}\\ &=\frac{e^{\pi i a/4}}{a}\sum _{n=0}^a (-1)^nf\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i n^2/a} \end{align}

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 $f\equiv 1$.

Edit: Here is a formal proof of the general fact using Poisson summation formula (see transition from the 3rd line to the 4th). \begin{align} \frac{1}{\sqrt{a}}\sum_{j=-\infty}^\infty a_je^{-\pi i j^2/a}&=\int_{-\infty}^\infty f(x) e^{\pi i a x^2}dx\\&=\int_{-1/2}^{1/2} f(x) \sum_{n=-\infty}^\infty e^{\pi i a (x+n)^2}dx\\ &=\sum_{n=-\infty}^\infty e^{\pi i a n}\int_{-1/2}^{1/2} f(x)e^{\pi i a x^2} e^{2\pi i a n x}dx\\ &=\frac{1}{a}\sum _{n=0}^a f\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i a\left(\frac{1}{2}-\frac{n}{a}\right)^2}\\ &=\frac{e^{\pi i (a-1)/4}}{a}\sum _{n=0}^a (-1)^nf\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i n^2/a} \end{align}

We will use the following well known fact (two proofs of this fact can be found at the end of the post):

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 $f\equiv 1$.

EDIT: 1st proof of the general fact using multisection. $$ \sum_{j\in\mathbb{Z}} a_je^{\pi i j^2/n}=\sum_{s=1}^ne^{\pi i s^2/n}\sum_{k\in\mathbb{Z}} a_{s+kn}e^{\pi i j^2/n}\\ =\sum_{s=1}^ne^{\pi i s^2/n}\frac{1}{n}\sum_{j=1}^ne^{2\pi isj/n}f(1/2-j/n)\\ =\frac{1}{n}\sum_{j=1}^nf(1/2-j/n)\sum_{s=1}^ne^{2\pi isj/n}e^{\pi i s^2/n} $$

The inner sum here is a Gauss sum and its value is known from other sources. The proof below avoids calculation of Gauss sums, but probably will be called unsatisfiable by some.

2nd proof of the general fact using Poisson summation formula (see transition from the 3rd line to the 4th).

\begin{align} \frac{1}{\sqrt{a}}\sum_{j=-\infty}^\infty a_je^{-\pi i j^2/a}&=\int_{-\infty}^\infty f(x) e^{\pi i a x^2}dx\\ &=\int_{-1/2}^{1/2} f(x) \sum_{n=-\infty}^\infty e^{\pi i a (x+n)^2}dx\\ &=\sum_{n=-\infty}^\infty e^{\pi i a n}\int_{-1/2}^{1/2} f(x)e^{\pi i a x^2} e^{2\pi i a n x}dx\\ &=\frac{1}{a}\sum _{n=0}^a f\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i a\left(\frac{1}{2}-\frac{n}{a}\right)^2}\\ &=\frac{e^{\pi i (a-1)/4}}{a}\sum _{n=0}^a (-1)^nf\left(\frac{1}{2}-\frac{n}{a}\right)e^{\pi i n^2/a} \end{align}