Here we give a proof that random walk is recurrent for 𝑝 =1/2 and not for other values of 𝑝.

From Shreve’s older book “Stochastic calculus and financial applications” p. 6 we have:
Let 𝑝 be the probability of a step to the right, 𝑋𝑖 =+1, and 𝑞 =1 −𝑝.
Let 𝑓⁡(𝑘) be the probability of reaching 𝐴 before −𝐵 given that 𝑆0 =𝑘. Then
𝑓⁡(𝑘)=𝑝⁢𝑓⁡(𝑘+1)+𝑞⁢𝑓⁡(𝑘−1)
In terms of Δ⁢𝑓⁡(𝑘) :=𝑓⁡(𝑘 +1) −𝑓⁡(𝑘) this says
Δ⁢𝑓⁡(𝑘)=(𝑞/𝑝)⁢Δ⁢𝑓⁡(𝑘−1)
so
Δ⁢𝑓⁡(𝑘+𝑗)=(𝑞/𝑝)𝑗⁢Δ⁢𝑓⁡(𝑘)
and so
𝑓⁡(𝑘)=𝑓⁡(−𝐵)+𝑘−1∑𝑖=−𝐵Δ⁢𝑓⁡(𝑖)=0+𝑘−1∑𝑖=−𝐵Δ⁢𝑓⁡(𝑖)=𝑘+𝐵−1∑𝑗=0Δ⁢𝑓⁡(𝑗−𝐵)
=Δ⁢𝑓⁡(−𝐵)⁢𝑘+𝐵−1∑𝑗=0(𝑞/𝑝)𝑗
=𝑓⁡(−𝐵+1)⁢𝑘+𝐵−1∑𝑗=0(𝑞/𝑝)𝑗
=𝑓⁡(−𝐵+1)⁢(𝑞/𝑝)𝑘+𝐵−1(𝑞/𝑝)−1
Since 𝑓⁡(𝐴) =1 we can evaluate 𝑓⁡(−𝐵 +1) as (𝑞/𝑝)−1(𝑞/𝑝)𝐴+𝐵−1.
Then
𝑓⁡(0)=(𝑞/𝑝)−1(𝑞/𝑝)𝐴+𝐵−1⁢(𝑞/𝑝)𝐵−1(𝑞/𝑝)−1
Thus

𝑃⁡(𝑆𝑛 hits 𝐴 before −𝐵∣𝑆0=0)=(𝑞/𝑝)𝐵−1(𝑞/𝑝)𝐴+𝐵−1
Setting 𝐴 =𝐵 we have
𝑃⁡(𝑆𝑛 hits 𝐴 before −𝐴∣𝑆0=0)=(𝑞/𝑝)𝐴−1(𝑞/𝑝)2⁢𝐴−1=1(𝑞/𝑝)𝐴+1
Let us consider the probability of never hitting −𝐴. This is the same as hitting 𝐴 before −𝐴, then 3⁢𝐴 before −𝐴 (so in effect 2⁢𝐴 before −2⁢𝐴), then 7⁢𝐴 before −𝐴 (so in effect 4⁢𝐴 before −4⁢𝐴), and so on, so we get
∏𝑘=1,2,4,…1(𝑞/𝑝)𝑘⁢𝐴+1=∞∏𝑛=01(𝑞/𝑝)𝐴⁢2𝑛+1
Let us set 𝐴 =2𝑎 where 𝑎 ≥0, 𝑎 an integer (where 𝑎 =0 is perhaps most interesting), so we are looking at
∞∏𝑛=01(𝑞/𝑝)2𝑛+𝑎+1
If 𝑝 ≤1/2 then 𝑞/𝑝 ≥1 and so this is
≤∞∏𝑛=0112𝑛+𝑎+1=0
since the factors are bounded below 1. But if 𝑝 >1/2 then 𝛼 :=𝑞/𝑝 <1 and we have the equivalent conditions ∞∏𝑛=01𝛼2𝑛+𝑎+1>0 log⁡∞∏𝑛=01𝛼2𝑛+1>−∞ ∑log⁡1𝛼2𝑛+1>−∞ ∑log⁡(𝛼2𝑛+1)<∞ But ln⁡(𝑥) ≤𝑥 −1 for all 𝑥 >0, so
∑log⁡(𝛼2𝑛+1)<∑𝛼2𝑛<∑𝛼𝑛<∞
Moreover as 𝑎 →∞ the nonzero product actually approaches 1. Thus the probability that each negative number will eventually be reached is zero when 𝑝 >1/2. Similarly, if 𝑝 <1/2 then the probability that each positive number will eventually be reached is zero.