Random walk on $\mathbb{Z}^d$ Problem Let $\{X_n\}_{n=0}^{\infty}$ be a random walk on $\mathbb{Z}^d$ such that; $X_0=(0,0,\cdots,0)$ and $\{X_n-X_{n-1}\}_{n=1}^{\infty}$ are mutually independent, identitically distributed $\mathbb{Z}^d$-valued random variables(not necessarily nearst neighbor). It's known that: $\exists \: C,c_1, c_2,\cdots, c_d \in \mathbb{R^+\cup{0}} \ \ \ \mbox{with} \ \ \sum_{i=1}^{d}c_i>1 \ \ \ \mbox{s.t.}$
$$ P((X_n)_i=0)\leq C n^{-c_i} \ \ \ \forall n\geq 1$$
Where $(X_n)_i$ is the $i$th coordinate of random variable $X_n$. 
Prove that the probability coming back to start point is less than $1$, ie. $P(\exists n\geq1 \ \ X_n=0)<1$

What I've done Since Fubini-Tonelli we have: 
$P(\exists n\geq1 \ \ X_n=0)<1 \Longleftrightarrow E [ \sum \limits_{n=0}^{\infty} \chi_{\{X_n=0\}}] < \infty$
$E [ \sum \limits_{n=0}^{\infty} \chi_{\{X_n=0\}}] =  \sum \limits_{n=0}^{\infty} E[\chi_{\{X_n=0\}}] = \sum \limits_{n=0}^{\infty} P(X_n=0) $  So we have:

$P(\exists n\geq1 \ \ X_n=0)<1 \Longleftrightarrow \sum \limits_{n=0}^{\infty} P(X_n=0)<\infty $
So we need a summable upper bound for $P(X_n=0)$. By conditional probability we know:
$P(X_n=0)=P((X_n)_1=0,(X_n)_2=0,\cdots,(X_n)_d=0)$
$=P((X_n)_1=0 \ |(X_n)_2=0,\cdots,(X_n)_d=0)\times \\ P((X_n)_2=0 \ |(X_n)_3=0,\cdots,(X_n)_d=0)\times \cdots \times \\ P((X_n)_{d-1}=0 \ |(X_n)_d=0) \times P((X_n)_d=0)$

Trouble: It would be great if following was true for $1\le i \le d-1$:
$P((X_n)_i=0 \ |(X_n)_{i+1}=0,\cdots,(X_n)_d=0) < P((X_n)_i=0)$
But is not. Actually right hand side might be also replaced by $K_n P((X_n)_i=0)$, for some suitable $K_n$. However I couldn't find. Could you please help me with this?
 A: Counterexample Let
$$S_n := \sum_{j=1}^n Y_j, \qquad n \in \mathbb{N} \tag{1}$$
a simple random walk on $\mathbb{Z}$, i.e. $Y_j \sim \frac{1}{2} (\delta_1+\delta_{-1})$ independent identically distributed random variables. By Stirling's formula, we have
$$\mathbb{P}(S_{2n}=0) = 2^{-2n} {2n \choose n} \sim \frac{1}{\sqrt{\pi n}} \qquad \text{and} \qquad \mathbb{P}(S_{2n+1}=0)=0 \tag{2}$$
for $n$ sufficiently large. Now we define a  random walk on $\mathbb{Z}^3$ by
$$X_n := ((X_n)_1,(X_n)_2,(X_n)_3)^T := \sum_{j=1}^n (Y_j,Y_j,Y_j)^T=(S_n,S_n,S_n)^T. $$
Then,
$$\mathbb{P}((X_n)_i=0) = \mathbb{P}(S_n=0) \stackrel{(2)}{\leq} \frac{C}{\sqrt{n}},$$
i.e. we can choose $c_i=1/2$ for $i=1,2,3$. Obviously, $\sum_{i=1}^3 c_i>1$. On the other hand, we have
$$\sum_{n=1}^{\infty} \mathbb{P}(X_n=0) = \sum_{n=1}^{\infty} \mathbb{P}((X_n)_1=0) = \sum_{n=1}^{\infty} \mathbb{P}(S_n=0) \stackrel{(2)}{=} \infty.$$
This means that $\mathbb{P}(X_n=0 \, \text{infinitely often})=1$ - and this contradicts the claim.
