Ah, I think I can stymy M with 2 nonconstant advisors. Namely, let A1(n)=12−1n+3 and A2(n)=12+1n+3. We (setting up an adversarial E) precommit to setting E(n)=0 if p(n)≥A2(n) and E(n)=1 if p(n)≤A1(n); now we can assume that M always chooses p(n)∈[A1(n),A2(n)], since this is better for M.

Now define b′i(j)=|Ai(j)+E(j)−1|−|p(j)+E(j)−1| and bi(n)=∑j<nb′i(j). Note that if we also define badi(n)=∑j<n(log|Ai(j)+E(j)−1|−log|p(j)+E(j)−1|) then ∑j<n|2bi(j)−badi(j)|≤∑j<n(2A1(j)−1−log(2A1(j))))=∑j<nO((12−A1(j))2) is bounded; therefore if we can force b1(n)→∞ or b2(n)→∞ then we win.

Let’s reparametrize by writing δ(n)=A2(n)−A1(n)=2n+3 and q(n)=p(n)−A1(n)δ(n), so that b′i(j)=δ(j)(|i−2+E(j)|−|q(j)−1+E(j)|).

Now, similarly to how M worked for constant advisors, let’s look at the problem in rounds: let s0=0, and sn=⌊exp(sn−1−1)⌋+1 for n>0. When determining E(sn−1),…,E(sn−1), we can look at p(sn−1),…,p(sn−1). Let tn=⌊b2(sn)−1n⌋. Let’s set E(sn−1),…,E(sn−1) to 1 if ∑sn−1j=sn−1δ(j)(1−q(j))≥1; otherwise we’ll do something more complicated, but maintain the constraint that b2(sn)≥b2(sn−1)−1n(n−1)≥tn−1+1n: this guarantees that tn is nondecreasing and that liminfj→∞b2(j)≥limn→∞tn.

If tn→∞ then b2(n)→∞ and we win. Otherwise, let t=limn→∞tn, and consider n such that tn−1=t.

We have ∑sn−1j=sn−1δ(j)(1−q(j))<1. Let J⊆{sn−1,…,sn−1} be a set of indices with q(j)≥q(j′) for all j∈J,j′∉J, that is maximal under the constraint that ∑j∈Jδ(j)(1−q(j))≤1n(n−1); thus we will still have ∑j∈Jδ(j)(1−q(j))≥1n(n−1)−δ(sn−1). We shall set E(j)=0 for all j∈J.

By the definition of J: ∑j∈Jb′1(j)=∑j∈Jδ(j)q(j)≥∑j∈Jδ(j)(1−q(j))∑sn−1j=sn−1δ(j)q(j)∑sn−1j=sn−1δ(j)(1−q(j))≥(1n(n−1)−δ(sn−1))∑sn−1j=sn−1δ(j)−11≥(1n(n−1)−δ(sn−1))(2log(sn+3sn−1+3)−1)≥2 if n≫0

For j′∉J, we’ll proceed iteratively, greedily minimizing ∣∣∑jj′=sn−11j′∉J(b′1(j′),b′2(j′))∣∣. Then: minsn−1≤j<snj∑j′=sn−11j′∉Jb′1(j′)≥−
⎷sn−1∑j=sn−1δ(j)2=−2
⎷sn−2∑j=sn−1+31j2≥−2
⎷sn−2∑j=sn−1+3(1j−1−1j)≥−2√sn−1+2≥−1 if n≫0

Keeping this constraint, we can flip (or not flip) all the E(j′)s for j′∉J so that ∑sn−1j′=sn−11j′∉Jb′2(j′)>0. Then, we have b2(sn)≥b2(sn−1)−1n(n−1), b1(sn)−b1(sn−1)=∑snj=sn−1(1j∈J+1j∉J)b′1(j)≥2−1=1 if n≫0, and for sn−1≤j≤sn, b1(j)≥b1(sn−1)+∑j−1j′=sn−11j′∉Jb′1(j′)≥b1(sn−1)−1 if n≫0.

Ah, I think I can stymy M with 2 nonconstant advisors. Namely, let A1(n)=12−1n+3 and A2(n)=12+1n+3. We (setting up an adversarial E) precommit to setting E(n)=0 if p(n)≥A2(n) and E(n)=1 if p(n)≤A1(n); now we can assume that M always chooses p(n)∈[A1(n),A2(n)], since this is better for M.

Now define b′i(j)=|Ai(j)+E(j)−1|−|p(j)+E(j)−1| and bi(n)=∑j<nb′i(j). Note that if we also define badi(n)=∑j<n(log|Ai(j)+E(j)−1|−log|p(j)+E(j)−1|) then ∑j<n|2bi(j)−badi(j)|≤∑j<n(2A1(j)−1−log(2A1(j))))=∑j<nO((12−A1(j))2) is bounded; therefore if we can force b1(n)→∞ or b2(n)→∞ then we win.

Let’s reparametrize by writing δ(n)=A2(n)−A1(n)=2n+3 and q(n)=p(n)−A1(n)δ(n), so that b′i(j)=δ(j)(|i−2+E(j)|−|q(j)−1+E(j)|).

Now, similarly to how M worked for constant advisors, let’s look at the problem in rounds: let s0=0, and sn=⌊exp(sn−1−1)⌋+1 for n>0. When determining E(sn−1),…,E(sn−1), we can look at p(sn−1),…,p(sn−1). Let tn=⌊b2(sn)−1n⌋. Let’s set E(sn−1),…,E(sn−1) to 1 if ∑sn−1j=sn−1δ(j)(1−q(j))≥1; otherwise we’ll do something more complicated, but maintain the constraint that b2(sn)≥b2(sn−1)−1n(n−1)≥tn−1+1n: this guarantees that tn is nondecreasing and that liminfj→∞b2(j)≥limn→∞tn.

If tn→∞ then b2(n)→∞ and we win. Otherwise, let t=limn→∞tn, and consider n such that tn−1=t.

We have ∑sn−1j=sn−1δ(j)(1−q(j))<1. Let J⊆{sn−1,…,sn−1} be a set of indices with q(j)≥q(j′) for all j∈J,j′∉J, that is maximal under the constraint that ∑j∈Jδ(j)(1−q(j))≤1n(n−1); thus we will still have ∑j∈Jδ(j)(1−q(j))≥1n(n−1)−δ(sn−1). We shall set E(j)=0 for all j∈J.

By the definition of J: ∑j∈Jb′1(j)=∑j∈Jδ(j)q(j)≥∑j∈Jδ(j)(1−q(j))∑sn−1j=sn−1δ(j)q(j)∑sn−1j=sn−1δ(j)(1−q(j))≥(1n(n−1)−δ(sn−1))∑sn−1j=sn−1δ(j)−11≥(1n(n−1)−δ(sn−1))(2log(sn+3sn−1+3)−1)≥2 if n≫0

For j′∉J, we’ll proceed iteratively, greedily minimizing ∣∣∑jj′=sn−11j′∉J(b′1(j′),b′2(j′))∣∣. Then: minsn−1≤j<snj∑j′=sn−11j′∉Jb′1(j′)≥− ⎷sn−1∑j=sn−1δ(j)2=−2 ⎷sn−2∑j=sn−1+31j2≥−2 ⎷sn−2∑j=sn−1+3(1j−1−1j)≥−2√sn−1+2≥−1 if n≫0

Keeping this constraint, we can flip (or not flip) all the E(j′)s for j′∉J so that ∑sn−1j′=sn−11j′∉Jb′2(j′)>0. Then, we have b2(sn)≥b2(sn−1)−1n(n−1), b1(sn)−b1(sn−1)=∑snj=sn−1(1j∈J+1j∉J)b′1(j)≥2−1=1 if n≫0, and for sn−1≤j≤sn, b1(j)≥b1(sn−1)+∑j−1j′=sn−11j′∉Jb′1(j′)≥b1(sn−1)−1 if n≫0.

Therefore, b1(j)→∞, so we win.