I would like to show that the following random walk on $\mathbb{Z}$ is transient. Starting at position $0$, we jump at a position at distance $n$ with probability equal to $\frac{|n|^{-1-\alpha}}{2\zeta(1+\alpha)}$, with $\alpha \in (0,1)$.
Formally, we define the random variable $X_k$ that denotes the position after $k$ steps as follow,
$$ \textrm{ if $n \neq 0$} \colon \quad \mathbb{P}\left[ X_{k+1} - X_k = n \right] = \frac{|n|^{-1-\alpha}}{2\zeta(1+\alpha)}, \quad \textrm{otherwise} \quad \mathbb{P}\left[ X_{k+1} - X_k = 0 \right] = 0. $$ The walk is transient if the integral $\int_{-\pi}^\pi \frac{1}{1- \varphi_X(t)} dt $ is finite, where $\varphi_X$ is the characteristic function. Writing down the characteristic function, I get that $$ \varphi_X(t) = \sum_{n = 1}^{\infty} \frac{|n|^{-1-\alpha}}{2\zeta(1+\alpha)}e^{itn} + \frac{|-n|^{-1-\alpha}}{2\zeta(1+\alpha)}e^{-itn} = \sum_{n = 1}^\infty \frac{n^{-1-\alpha}}{\zeta(1+\alpha)} \cos(tn)$$ So the goal is to estimate this last sum. In particular, I would like to show that the sum is less than $1$ for any $t \neq 0$.
My intuition was to use a Taylor expansion of $\cos$ near $0$ i.e., $\cos(x) \simeq 1 - O\left(\frac{x^2}{2}\right)$ but I don't know how to proceed next.
Here are some details which may be helpful. As per my comment, they did not come out quite as I desired, so it is not a complete solution by any means.
It is immediate that $\lvert \varphi_X(t) \rvert < 1$ for $t \notin 2 \pi \mathbb Z$. Indeed, without the $\cos$ term it is precisely 1, and for any $t \notin 2 \pi \mathbb Z$ we have $\lvert \cos(t n) \rvert \ne 1$ for many $n \in \mathbb N$. This doesn't stop the possibility of $\varphi_X(t) \to 1$ as $t \to 0$; it is this rate which must be controlled.
For all $\epsilon > 0$, there exists a constant $N = N_\epsilon$ so that $$ \textstyle \bigl\lvert \varphi(t) - \sum_{n=1}^N n^{-1-\alpha} \cos(tn)/\zeta(1+\alpha) \bigr\rvert \le \epsilon \quad\text{for all}\quad t \in [-\pi, \pi]. $$ Indeed, something like $N_\epsilon := 1/\epsilon$ will work for $\epsilon$ small enough. By the same argument, replacing $N$ by something larger if necessary, writing $\psi(t) := 1 - \varphi(t)$, we have $$ \textstyle \bigl\lvert \psi(t) - \sum_{n=1}^N \bigl( 1 - \cos(tn) \bigr) n^{-1-\alpha} / \zeta(1+\alpha) \bigr\lvert \le \epsilon \quad\text{for all}\quad t \in [-\pi,\pi]. $$ Now replace $1 - \cos(tn)$ with $\tfrac12(tn)^2 + \mathcal O\bigl( (tn)^3 \bigr)$. For $\eta > 0$ sufficiently small (depending on $N$, and so on $\epsilon$), since $\psi$ is uniformly bounded, we obtain $$ \textstyle \bigl\lvert \psi(t) - \sum_{n=1}^N \tfrac12 (tn)^2 n^{-1-\alpha} / \zeta(1+\alpha) \bigr\lvert \le 2 \epsilon \quad\text{for all}\quad t \in [-\eta, \eta].$$ So we see that the leading order is $t^2$. Note that $1/t^2$ is not integrable.