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.



Parameterized simple asymmetric random walk
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](https://link.springer.com/book/10.1007/978-3-642-65809-9) (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 $Y\_n:=X\_n-EX\_n$.


