Parameterized simple asymmetric random walk Let $ t>0 $, and we look at the random walk $S_{n}=\sum_{i=1}^{n}X_{n}$ on $\mathbb{Z}$ with $S_0=0$ where $$ \mathbb{P}\left(X_{n}=1\right)    =\frac{1}{2}\left(1+\frac{1}{n^{t}}\right)
$$ $$ \mathbb{P}\left(X_{n}=-1\right)   =1-\mathbb{P}\left(X_{n}=1\right)=\frac{1}{2}\left(1-\frac{1}{n^{t}}\right)$$We would like to know when this walk is recurrent and when it is transient, depending on $t$. $S_{n}$ is not a martingale, and we have $ \mathbb{E}\left(X_{n}\right)=\frac{1}{n^{t}}$. so we have $ \mathbb{E}\left(S_{n}\right)=\sum_{i=1}^{n}\frac{1}{i^{t}}$ which might look look like something we can apply series convergence to, but I didn't achieve much with this approach.
I tried to bound  $ \mathbb{P}\left(S_{2n}=0\right)$: $$ \mathbb{P}\left(S_{2n}=0\right)    =\sum_{s\in S}\left(\prod_{i=1}^{2n}\mathbb{P}\left(X_{i}=s_{i}\right)\right)
    \leq{2n \choose n}\left(\prod_{i=1}^{n}\mathbb{P}\left(X_{i}=1\right)\right)\left(\prod_{i=n+1}^{2n}\mathbb{P}\left(X_{i}=-1\right)\right)$$
(Where $ S $  is the set of all arrangements of $n$ $-1$s and $n$ $1$s) But this seems like a pretty bad bound, since the sum of it diverges to $\infty $. We also know that $ \mathbb{P}\left(X_{1}=1\right)=1$, so we can change ${2n \choose n}$ to ${2n-1 \choose n}$, but this doesn't achieve much either. (and anyway, the above bound isn't that easy to work with)
My current attempt is to bound $ \mathbb{P}\left(S_{2n}=0\right)$ from above with $ \mathbb{P}\left(S_{2n}=0\right) \leq p_n$, with $\sum_{n=1}^{\infty}p_{n}<\infty$  which implies $\sum_{i=1}^{\infty}\mathbb{P}\left(S_{2i}=0\right)<\infty$ and then using Borel–Cantelli lemma we will get $\mathbb{P}\left(S_{2i}=0\quad\text{i.o.}\right)=0$. But all of my attempts to get a bound which is (1) convergent (2) easy to work with, have failed.
 A: Recall that a random walk (or a Markov chain in general) is called recurrent if it almost surely (a.s.) returns to the initial state infinitely often.
We will show that in our case the walk is recurrent iff $t\ge1/2$.
The key here is the law of the iterated logarithm. Indeed, according to (say) Theorem 1 of Chapter X (see the statement of this theorem reproduced at the end of this answer),
\begin{equation*}
\begin{aligned}
    &\limsup_n\frac{S_n-ES_n}{\sqrt{2n\ln\ln n}}=1,\\ 
    &\liminf_n\frac{S_n-ES_n}{\sqrt{2n\ln\ln n}}=-1 
\end{aligned}
\tag{1}\label{1}
\end{equation*}
a.s.
Also, if $t\ge1/2$, then
\begin{equation*}
    ES_n=\sum_{i=1}^n EX_i=\sum_{i=1}^n 1/i^t=O(n^{1/2})=o(\sqrt{2n\ln\ln n}), 
\end{equation*}
(as $n\to\infty$), so that, by \eqref{1},
\begin{equation*}
\begin{aligned}
    &\limsup_n\frac{S_n}{\sqrt{2n\ln\ln n}}=1,\\ 
    &\liminf_n\frac{S_n}{\sqrt{2n\ln\ln n}}=-1  
\end{aligned} 
\end{equation*}
a.s., which implies that the walk is recurrent (because it cannot jump over $0$: If $S_k<0<S_m$ for some natural $k$ and $m$, then $S_j=0$ for some natural $j$ between $k$ and $m$).
If now $t<1/2$, then
\begin{equation*}
    ES_n=\sum_{i=1}^n EX_i=\sum_{i=1}^n 1/i^t\asymp n^{1-t},
\end{equation*}
so that, by \eqref{1}, $S_n-ES_n=o(ES_n)$ a.s. and hence $S_n\sim ES_n\to\infty$ a.s., which implies that the walk is not recurrent.
Thus, the walk is recurrent iff $t\ge1/2$. (Basically, it all depends on how $n^{1-t}$ asymptotically compares with $\sqrt{n\ln\ln n}$)

The formulation of the law of the iterated logarithm used in this answer:

Let $Y_1,Y_2,\dots$ be independent zero-mean random variables with $s_k^2:=EY_k^2<\infty$ for all $k$. Let $T_n:=\sum_{k=1}^n Y_k$ and $B_n:=\sum_{k=1}^n s_k^2$. Suppose that $B_n\to\infty$ and for some sequence $(M_n)$ of positive constants we have $|Y_n|\le M_n$ for all $n$ and
\begin{equation*}
    M_n=o\Big(\sqrt{\frac{B_n}{\ln\ln B_n}}\Big). 
\end{equation*}
Then
\begin{equation*}
\begin{aligned}
    &\limsup_n\frac{T_n}{\sqrt{2B_n\ln\ln B_n}}=1
\end{aligned}
\end{equation*}
a.s.

We used this result for $Y_n:=X_n-EX_n$.
A: Recall that a random walk (or a Markov chain in general) is called recurrent if it almost surely (a.s.) returns to the initial state infinitely often.
The previous answer, which showed that the walk (in question) is recurrent iff $t\ge1/2$, was based on a direct application of the law of the iterated logarithm, which is in turn based on certain upper and lower bounds on probabilities of large (or, rather, so-called moderate) deviations of sums of independent random variables.
The "My current attempt [...]" edit and a comment by the OP subsequent to the previous answer seem to suggest that the OP prefers an indirect argument, based on the following criterion (see e.g. Theorem 13.1):

The random walk will be recurrent iff
\begin{equation*}
    \sum_n P(S_{2n}=0)=\infty. \tag{2}\label{2}
\end{equation*}

To estimate the probabilities $P(S_{2n}=0)$ accurately enough, we need an appropriate local (central) limit theorem. Let $Y_j:=(1-X_j)/2$, so that $Y_j$ takes values in the set $\{0,1\}$ and $P(Y_j=0)=\frac12\,(1+\frac1{j^{t}})>\frac12\,(1-\frac1{j^{t}})=P(Y_j=1)$. Let $T_{2n}:=\sum_{j=1}^{2n}Y_j=n-S_n/2$. Applying now Theorem 4 of Chapter VII (with the $Y_j$'s in place of the $X_j$'s), we get
\begin{equation*}
    P(S_{2n}=0)=P(T_{2n}=n)=\frac1{\sqrt{2\pi B_{2n}}}\,\exp\Big(-\frac{(n-M_{2n})^2}{2B_{2n}}\Big)
    +O(1/B_{2n}), 
\end{equation*}
where $B_{2n}:=\sum_{j=1}^{2n}E(Y_j-EY_j)^2\sim n/2$ and $M_{2n}:=\sum_{j=1}^{2n}EY_j=n-ES_{2n}/2$.
Consider first the case $t\ge1/2$. Then
\begin{equation*}
    0\le ES_{2n}=\sum_{i=1}^{2n} EX_i=\sum_{i=1}^{2n} 1/i^t\le\sum_{i=1}^{2n} 1/i^{1/2}\sim2\sqrt{2n}
\end{equation*}
(as $n\to\infty$). So, for some $u_n\in[0,1]$,
\begin{equation*}
    P(S_{2n}=0)=\frac{1+o(1)}{\sqrt{\pi n}}\,e^{-2u_n}
    +O(1/n)\asymp\frac1{\sqrt n}. 
\end{equation*}
So, if $t\ge1/2$, then \eqref{2} holds and hence the random walk is recurrent.
It remains to consider the case $t<1/2$. Then write
\begin{equation*}
    P(S_{2n}=0)\le P(S_{2n}\le0)=P(S_{2n}-ES_{2n}\le-ES_{2n}) 
    \le\frac{E|(S_{2n}-ES_{2n}|^p}{(ES_{2n})^p}
\end{equation*}
for real $p>0$. If $p\ge2$, then, by Rosenthal's inequality, $E|S_{2n}-ES_{2n}|^p=O(n^{p/2})$. Letting now $p=\dfrac2{1/2-t}$, we get
\begin{equation*}
    P(S_{2n}=0)=O\Big(\frac{n^{p/2}}{n^{(1-t)p}}\Big)=O\Big(\frac1{n^2}\Big). 
\end{equation*}
So, if $t<1/2$, then \eqref{2} does not hold and hence the random walk is not recurrent.
Thus, indeed the walk is recurrent iff $t\ge1/2$. $\quad\Box$
A: $\newcommand\ep\epsilon$The OP, who wanted more elementary arguments, agreed that the cases $t<1/2$ and $t>1$ would be enough. Therefore I am providing the third answer to the question, with more elementary arguments for $t<1/2$ and for $t>1$.
Consider first the case $t<1/2$. We have
\begin{equation*}
    ES_{2n}=\sum_{i=1}^{2n} EX_i=\sum_{i=1}^{2n} 1/i^t\ge c n^{1-t}  
\end{equation*}
for some real $c>0$ and all $n$.
So, by Hoeffding's inequality (see the second displayed inequality, with $a_i=-1$, $b_i=1$, $t=ES_{2n}$),
\begin{equation}
    P(S_{2n}=0)\le P(S_{2n}\le0)=P(S_{2n}-ES_{2n}\le-S_{2n}) \\ 
    \le2\exp\Big(-\frac{2c^2n^{2(1-t)}}{4n}\Big)
    =2e^{-(c^2/2)n^{1-2t}}=O(1/n^2). 
\end{equation}
So,
\begin{equation*}
    \sum_n P(S_{2n}=0)<\infty 
\end{equation*}
and hence the walk is not recurrent, if $t<1/2$.
Consider now the case $t>1$. Let $U_1,U_2,\dots$ be independent random variables (r.v.'s) each uniformly distributed on the interval $[0,1]$. Without loss of generality, $X_j=1(U_j<\frac12\,(1+\frac1{j^t}))-1(U_j>\frac12\,(1+\frac1{j^t}))$ for all natural $j$.
Let $Y_j=1(U_j<\frac12)-1(U_j>\frac12)$ for all natural $j$, so that
the $Y_j$'s are independent r.v.'s with $P(Y_j=1)=P(Y_j=-1)=1/2$ for all $j$. Let
$T_k:=\sum_{j=1}^k Y_j$.
Take any real $\ep>0$. Then
\begin{equation}
    p_n:=P(\exists j\ge2n\ X_j\ne Y_j)\le\ep
\end{equation}
for some natural $n=n_\ep$, because
\begin{equation}
    p_n\le\sum_{j\ge 2n}P(X_j\ne Y_j)=\sum_{j\ge 2n}\frac1{2j^t}\to0
\end{equation}
as $n\to\infty$.
Let "i.o." mean "for infinitely many natural $m$ such that $m\ge n$". Since the Markov chain corresponding to the $Y_j$'s is irreducible and recurrent, we have
\begin{equation}
\begin{aligned}
    1&=P(T_{2m}=T_{2n}-S_{2n}\ i.o.) \\ 
    &=  P(T_{2m}-T_{2n}+S_{2n}=0\ i.o.) \\ 
    &\le P(S_{2m}-S_{2n}+S_{2n}=0\ i.o.)+P(\exists m\ge n\ T_{2m}-T_{2n}\ne S_{2m}-S_{2n} ) \\ 
    &\le P(S_{2m}-S_{2n}+S_{2n}=0\ i.o.)+p_n \\ 
    &=P(S_{2m}=0\ i.o.)+p_n \\ 
    &\le P(S_{2m}=0\ i.o.)+\ep,
\end{aligned}
\end{equation}
so that $P(S_{2m}=0\ i.o.)\ge1-\ep$, for each $\ep>0$. So, $P(S_{2m}=0\ i.o.)=1$. That is, the walk is recurrent, if $t>1$. $\quad\Box$
