For asymmetric one-dimensional simple random walk, that is $$P(X_n = X_{n-1} + 1) = p = 1 - P(X_n = X_{n-1} - 1)$$ for some $p \ne 1/2$, provide an asymptotic upper bound for $\sum_{n=N}^{\infty}p_{ii}^{(n)}.$ We need to find an explicit expression $\alpha_N$ with the property $\lim_{N \to \infty} \alpha_{N} = 0$, such that $\sum_{n=N}^{\infty} p_{ii}^{(n)} \le \alpha_{N}$ for all large enough N.
Attempt:
Since this random walk is different at $p = \frac{1}{2}$, we know that $E(X_n|p= \frac{1}{2}) = 0$. If $p \ne \frac{1}{2}$ then since $X_n$ is Bernoulli random variable, $E(X_n) = np,$ hence $Var(X_n) = np(1-p).$ Therefore we apply the central limit theorem, the distribution of $X_n$ will be tend to the normal distribution. I suspect that if we simplify $\sum_{n=N}^{\infty} p_{ii}^{(n)}$ as convergent series, then we should be able to define $\alpha_{n}.$ Since in this random walk we move either 1 unit to the right or left, $p_{ii} = 0$, not sure how to determine $p_{ii}^{(n)}$.
$$\sum_{n=2N}^{\infty}p_{ii}^{(n)} = \sum_{n=N}^\infty {2n \choose n} p^n (1-p)^n$$
Via the Stirling formula: $${2n \choose n}\sim \frac{2^{2n}}{\sqrt{\pi n}}$$ so $${2n \choose n} p^n (1-p)^n \sim \frac{(4p(1-p))^{n}}{\sqrt{\pi n}}$$ As $p\neq \frac 12$, $0\le 4p(1-p)<1$ and the series
$$\sum {2n \choose n} p^n (1-p)^n$$ and $$\sum\frac{(4p(1-p))^{n}}{\sqrt{\pi n}}$$ are both convergent, and their rest are equivalent. So you can choose $$ \alpha_N = 2\sum_{[N/2]}^\infty\frac{(4p(1-p))^{n}}{\sqrt{\pi n}}\to0 $$ (the 2 is here to ensure you have the inequality for $N$ large enough).