[{"type":"thinking","thinking":"Committed. Next: summability lemma + finite cutoff. In scratch:\n\n```lean\n-- the term family on the subtype\nnoncomputable def sTerm (n : ℕ) (m : {m : ℕ // Odd m ∧ 3 ≤ m}) : ℝ :=\n  (1 - 1 / ((m : ℕ) : ℝ)) ^ n - 1 + (n : ℝ) / (m : ℕ)\n\ntheorem sTerm_summable (n : ℕ) : Summable (sTerm n) := by\n  -- comparison with C/m²\n  have hC : Summable (fun m : {m : ℕ // Odd m ∧ 3 ≤ m} => (n : ℝ)*(n-1)/(2 * ((m:ℕ):ℝ)^2)) := by\n    have h := (summable_one_div_nat_pow (p := 2)).mpr (by norm_num)\n    -- h : Summable (fun n => 1/(n:ℝ)^2)\n    have h2 := h.mul_left ((n:ℝ)*(n-1)/2)\n    -- h2 : Summable (fun m => C * (1/m²))\n    -- need subtype version: Summable.subtype\n    have h3 := h2.subtype (fun m : ℕ => Odd m ∧ 3 ≤ m)\n    -- h3 : Summable (fun m : subtype => C * (1/↑m²))? then congr to C/(2?)... forms\n    ...\n  apply Summable.of_nonneg_of_le (fun m => bern_term_nonneg n m m.property.2.trans?...) ...\n```\n\nCareful details:\n- m.property : Odd ↑m ∧ 3 ≤ ↑m, so m ≥ 3 → 1 ≤ m ✓.\n- summable_one_div_nat_pow gives `Summable (fun n => 1 / (n:ℝ)^p)` — at n=0: 1/0 = 0 ✓ summable handles.\n- mul_left C: Summable (fun m => C * (1/(m:ℝ)^2)) — then subtype: `Summable.subtype (hf) (p)` — signature from Multipliable.subtype (hf) (p : β → Prop): need to check the resulting function form: Summable (fun x : Subtype p => f ↑x)? The tsum function on subtype in SInfty is `fun m : {m // Odd m ∧ 3 ≤ m} => ((1 - 1/(m:ℝ))^n - 1 + n/m)` where (m : ℝ) coercion is ↑↑m — `↑(m:Subtype) : ℕ` then to ℝ? In SInfty def I wrote `(m : ℝ)` with m : {m : ℕ // ...} — that elaborates as ((↑m : ℕ) : ℝ) via double coercion ✓.\n- of_nonneg_of_le (hg : ∀ b, 0 ≤ g b) (h : ∀ b, g b ≤ f b) (hf : Summable f) : Summable g — check name: `Summable.of_nonneg_of_le` ✓ exists (Order.lean).\n- The comparison f = fun m : subtype => C * (1/(↑↑m)^2) vs g ≤ f pointwise needs bern_term_upper's conclusion ≤ n(n−1)/(2m²) = C*(1/m²) with C = n(n−1)/2 — same thing by ring. Use congr/funext to massage.\n\n- Then finite cutoff: Finset s_K := (Finset.Icc 1 K).image (fun k => ⟨2*k+1, by ...⟩ : {m // Odd m ∧ 3 ≤ m}) — the image needs a function into the subtype with proofs — dependent image: `Finset.image` needs the function; proving membership in subtype inside the lambda — use `Finset.attach`? Alternative cleaner: use Finset.map with an embedding k ↦ ⟨2k+1, proof⟩: `Function.Embedding` with proof fields — proofs are Props, subsingleton — define:\n```lean\ndef oddEmbedding : ℕ ↪ {m : ℕ // Odd m ∧ 3 ≤ m} where\n  toFun k := ⟨2*k+3, by constructor; · exact ⟨k+1, by ring⟩; · omega⟩  -- Odd (2k+3): Odd m means ∃ k, m = 2k+1: 2k+3 = 2(k+1)+1 ✓\n  inj' := by intro a b h; ... (Subtype.ext_iff → 2a+3 = 2b+3 → a = b by omega/linarith)\n```\nHmm — but my sum range: m = 2k+1 for k ∈ Icc 1 K ↔ m = 2j+3 for j ∈ range... use j = k−1, j ∈ Finset.range K (k = j+1 ∈ Icc 1 K ✓). So embedding j ↦ 2j+3, Finset.range K mapped.\n\nAlternatively avoid subtypes in the finite sum: use `sum_le_tsum` with Finset of the subtype via `Finset.univ` tricks... no. Direct approach:\n```lean\nhave cutoff : ∑ k ∈ Finset.Icc 1 K, ((1 - 1/((2*k+1:ℕ):ℝ))^n - 1 + (n:ℝ)/(2*k+1)) ≤ SInfty's tsum...\n```\nRoute: `Summable.sum_le_tsum (s : Finset α) (hf : Summable f) (hg : ∀ i ∉ s, 0 ≤ f i)` : ∑ i in s, f i ≤ ∑' i, f i. With α = subtype, s = image of embedding. Then rewrite the sum via Finset.sum_map / sum_image + injectivity:\n∑ m ∈ s.image emb, f m = ∑ j ∈ range K, f (emb j) — `Finset.sum_map (s) (e) (f)` ✓ (map with embedding). f (emb j) = ((1 − 1/((2j+3:ℕ):ℝ))^n − 1 + n/(2j+3)).\n\nThen per-term drop: f(emb j) ≥ n/(2j+3) − 1 since (1−1/m)ⁿ ≥ 0 (pow_nonneg: 1 − 1/m ≥ 0 for m ≥ 1 ✓).\n\nSo tsum ≥ Σ_{j<K} (n/(2j+3) − 1) = n·Σ_{j<K} 1/(2j+3) − K.\n\nThen odd harmonic identity: Σ_{j<K} 1/(2j+3 : ℝ) = Σ_{k=1}^K 1/(2k+1) (k=j+1 ✓ range vs Icc shift: Finset.sum_range_succ' or sum_Icc_eq_sum_range) = H_{2K+1} − ½H_K − 1 (induction).\n\nThis is all going fine but long. Given the session length so far and natural break point, let me consider pacing: we've been going a while. The user's directive: break prompt near 5 PM; it's later now (~20:30?). Actually I don't know the current time — session started earlier; builds suggest evening. Let me check time and consider asking the user about a break. The instructions: \"BREAK PROMPT: if work is running long — ESPECIALLY approaching 5:00 PM local time — pause and ASK the user\". Let me check the clock.","signature":""},{"type":"text","text":"Two more lemmas proved and committed. This is a natural checkpoint — let me check the time before continuing deeper into the summability chain."},{"type":"tool_use","id":"Bash_211","name":"Bash","input":{"command":"date \"+%H:%M %Z\"","description":"Check local time"},"extra_content":null}]