= Solution
A state is recurrent if, starting there, the probability of returning at some strictly positive time is one. It is transient if that return probability is strictly less than one. Suppose $i,j$ communicate. Choose $r,s$ with $p_{ji}(r)>0$ and $p_{ij}(s)>0$. Chapman–Kolmogorov gives
$$
p_{jj}(r+n+s)\ge p_{ji}(r)p_{ii}(n)p_{ij}(s).
$$
If $i$ is recurrent, summing over $n$ and using the supplied return-series criterion makes the return series for $j$ diverge, so $j$ is recurrent. Interchanging the states proves equivalence; therefore transience is also constant on a <communicating class>.
For the nearest-neighbor walk, return in an odd number of steps is impossible. A return in $2n$ steps requires exactly $n$ steps each way, so
$$
p_{ii}(2n)=\binom{2n}{n}[p(1-p)]^n\sim\frac{[4p(1-p)]^n}{\sqrt{\pi n}}.
$$
At $p=1/2$, the return series diverges like $\sum n^{-1/2}$. If $0<p<1$ and $p\ne1/2$, then $4p(1-p)<1$, so it converges by exponential decay. At $p=0$ or one the walk never returns. Hence $\boxed{\text{recurrence occurs exactly at }p=1/2}$.
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