[{"type":"thinking","thinking":"`Summable.comp_injective (hf : Summable f) (hi : Injective i) : Summable (f ∘ i)` exists (Group.lean:294 area, the to_additive version). And line 303 shows `hf.comp_injective Subtype.coe_injective` — exactly the pattern for subtypes.\n\nSo plan:\n- f : ℕ → ℝ := fun m => if 2 ≤ m then C/(m:ℝ)² else 0 — hmm, actually simpler: f m = C * (1/(m:ℝ)²) is summable directly? 1/(m:ℝ)² at m=0 is 1/0 = 0 in Lean (division by zero = 0), so `fun m => C * (1/(m:ℝ)^2)` is summable: `(summable_one_div_nat_pow.mpr (by norm_num : 1 < 2)).mul_left C`. ✓ And term(m) ≤ C/m² = C * (1/m²) ✓ for all m ≥ 1. Nonneg ✓.\n\n- Subtype summability: `Summable.of_nonneg_of_le (fun m => term_nonneg) (fun m => term_le) ((p-series).comp_injective Subtype.coe_injective)` — wait, but the comparison on the subtype: `Summable.of_nonneg_of_le (hg : ∀ b, 0 ≤ g b) (hgb : ∀ b, g b ≤ f b) (hf : Summable f)` where f is the comparison family on the subtype. So f := (fun m : ℕ => C/(m:ℝ)²) ∘ (↑). Summable via comp_injective ✓. Then `Summable.of_nonneg_of_le` gives Summable (fun m : subtype => term m) ✓.\n\nThen `Summable.sum_le_tsum s (fun i _ => term_nonneg i) hsummable` gives `∑ i ∈ s, term i ≤ ∑' i, term i` where s : Finset subtype. ✓\n\nThe finset s = image of k ↦ ⟨2k+1, odd, ≥3⟩ over Icc 1 K. Sum over image with injectivity: `Finset.sum_image` needs proof the map is injective on the set. Map: k ↦ (⟨2*k+1, h1, h2⟩ : subtype). Injective because Subtype.val ∘ map = 2k+1 injective.\n\n`Finset.sum_image (h : ∀ x ∈ s, ∀ y ∈ s, g x = g y → x = y) : ∑ x ∈ s.image g, f x = ∑ x ∈ s, f (g x)`.\n\nThen each term f(g k) = term(2k+1) ≥ n/(2k+1) − 1 by Bernoulli.\n\nSum manipulation: `∑ k ∈ Icc 1 K, (n/(2k+1) − 1) = n * (∑ k ∈ Icc 1 K, 1/(2k+1)) − K`. Cardinality of Icc 1 K = K (for K ≥ 1). Use `Finset.sum_sub_distrib`, `Finset.mul_sum`, `Nat.card_Icc` → `Finset.card_Icc : card (Icc a b) = b + 1 − a`, card = K+1−1 = K ✓.\n\nOdd-harmonic identity by induction — but with harmonic being ℚ-valued cast to ℝ, sums over my real A_K. Actually I don't need to relate A_K to Mathlib's `harmonic` via my own sum — I can prove directly:\n\n`harmonicOdd (K : ℕ) : ℝ := ∑ k ∈ Finset.Icc 1 K, (1 : ℝ)/(2*k+1)`\nLemma: `harmonicOdd K = (harmonic (2*K+1) : ℝ) − (harmonic K : ℝ)/2 − 1`.\n\nInduction on K. K=0: LHS empty sum = 0; RHS: harmonic 1 = 1, harmonic 0 = 0 → 1 − 0 − 1 = 0 ✓. Step K→K+1: Icc 1 (K+1) = insert (K+1) (Icc 1 K) for... `Finset.sum_Icc_succ_top (h : a ≤ b + 1) : ∑ k ∈ Icc a (b+1), f k = ∑ k ∈ Icc a b, f k + f (b+1)`. ✓\n\nRHS step: harmonic(2(K+1)+1) = harmonic(2K+3) = harmonic(2K+1) + 1/(2K+2) + 1/(2K+3) (two harmonic_succ steps, casts via push_cast/Rat.cast lemmas). harmonic(K+1)/2 = harmonic(K)/2 + 1/(2(K+1)). And 1/(2K+2) = (1/2)·(1/(K+1)) ✓ field.\n\nThen harmonic bounds:\n- H1: for M ≥ 1: log(M:ℝ) + γ ≤ (harmonic M : ℝ). From `eulerMascheroniConstant_lt_eulerMascheroniSeq' M` — need to unfold eulerMascheroniSeq' at M ≠ 0: `eulerMascheroniSeq' M = (harmonic M : ℝ) − log M` when M ≠ 0 (if_false). The lemma gives strict <. Use .le.\n- H2: (harmonic M : ℝ) ≤ log(M+1) + γ from `eulerMascheroniSeq_lt_eulerMascheroniConstant M`: eulerMascheroniSeq M = harmonic M − log(M+1) < γ. Note: in the def, `log (n + 1)` with (n : ℕ) cast — `log (↑n + 1)`. I need log(K+1) where K : ℕ — matches with push_cast: ((K:ℝ)+1) = ((K+1 : ℕ) : ℝ) ✓.\n\nTerm bound details. Define within proof: for m : ℕ, 1 ≤ m:\n- bernoulli: `1 + n*(-1/m) ≤ (1 + (-1/m))^n` i.e. `1 − n/m ≤ (1 − 1/m)^n` from `one_add_mul_le_pow (a := -(1/m)) (by positivity-ish : -2 ≤ -1/m) n`. −2 ≤ −1/m ⟺ 1/m ≤ 2 ✓ for m ≥ 1: 1/m ≤ 1 ≤ 2. \n\n- term_nonneg: (1−1/m)^n − 1 + n/m ≥ 0 ⟸ (1−1/m)^n ≥ 1 − n/m ✓.\n\n- term_lower: term ≥ n/m − 1: same Bernoulli.\n\n- term_upper: term ≤ C(n,2)/m² where C(n,2) = n(n−1)/2 as real: ((n:ℝ)*((n:ℝ)−1)/2) — avoid Nat.choose casts; just use n(n−1)/2 real.\n\nProof of term_upper: Let y = 1 − 1/m. Note 1 − y = 1/m.\n1 − y^n = (1−y) * Σ_{j ∈ range n} y^j: from `geom_sum_mul y n : (∑ j ∈ range n, y^j) * (y − 1) = y^n − 1`. So 1 − y^n = (1 − y) * Σ y^j by neg: (Σ y^j)*(1−y) = 1 − y^n ✓ (multiply by −1).\nterm = y^n − 1 + n/m = n/m − (1 − y^n) = n*(1/m) − (1/m)*Σ y^j = (1/m)*(n − Σ_{j<n} y^j) = (1/m)*Σ_{j<n} (1 − y^j).\nEach: 1 − y^j ≤ j/m ⟸ y^j ≥ 1 − j*(1/m) (Bernoulli with j) ✓ and 1 − y^j ≥ 0 ⟸ y^j ≤ 1 (0 ≤ y ≤ 1: y = 1 − 1/m ∈ [0,1) for m ≥ 1; pow_le_one). \nΣ_{j<n} (1 − y^j) ≤ Σ_{j<n} j/m = (1/m) Σ_{j<n} j = (1/m) * (n(n−1)/2): `Finset.sum_range_id`? `∑ i ∈ range n, i = n(n−1)/2` — `Finset.sum_range_id : ∑ i ∈ range n, i = n * (n − 1) / 2` (ℕ division!). Casts of ℕ division are painful. Work over ℝ/ℚ: `Finset.sum_range_id_mul_two : (∑ i ∈ range n, i) * 2 = n * (n − 1)` — that's the ℕ version; casting that equality gives ((∑ i, (i:ℝ)) * 2 = n*(n−1)) ✓ then divide. \n\nSo term ≤ (1/m)*(1/m)*(n(n−1)/2) = (n(n−1)/2)/m² ✓. Need m > 0 to manipulate 1/m: m ≥ 1 ✓ (subtype has m ≥ 3; but prove bounds for m ≥ 1 for reuse).\n\nActually the nonneg side for of_nonneg_of_le: on subtype elements m ≥ 3 ≥ 1 ✓.\n\nNow assembly (the main theorem). Sketch in Lean:\n\n```lean\ntheorem sInfty_lower_bound (n : ℕ) (hn : 20 ≤ n) :\n    (1/2) * n * Real.log n + (lagariasC1 - γ/2) * n - 4 ≤ SInfty n := by\n  -- abbreviations\n  set γ := Real.eulerMascheroniConstant\n  -- term function\n  set g : ℕ → ℝ := fun m => (1 - 1/(m:ℝ))^n - 1 + n/(m:ℝ)\n  -- bounds on g for m ≥ 1\n  have g_nonneg : ∀ m : ℕ, 1 ≤ m → 0 ≤ g m\n  have g_le : ∀ m : ℕ, 1 ≤ m → g m ≤ ((n:ℝ)*((n:ℝ)-1)/2) * (1/(m:ℝ)^2)\n  have g_ge : ∀ m : ℕ, 1 ≤ m → g m ≥ (n:ℝ)/(m:ℝ) - 1\n  -- summability\n  have hsumm : Summable fun m : {m : ℕ // Odd m ∧ 3 ≤ m} => g m\n  -- partial sum ≤ tsum\n  obtain ⟨q, hq⟩ : ∃ q, n = 2*q ∨ n = 2*q+1 := ...  -- from even_or_odd\n  set K := if Even n then q-1 else q  -- handle by cases instead\n  ...\n```\n\nHandling two cases: rcases on n even/odd:\n- `Nat.even_or_odd n : Even n ∨ Odd n`. `Even n` unfolds to `∃ r, n = r + r`; `Odd n` to `∃ k, n = 2*k + 1`.\n\nCase even: obtain ⟨q, rfl⟩; n = q+q. Have 20 ≤ q+q so q ≥ 10. K := q−1. Facts: 2*K+1 = 2q−1 = n−1 (nat, q ≥ 1); K+1 = q; (K:ℝ) = q − 1 etc.\n\nThe finset image: s := (Finset.Icc 1 K).image (fun k => ⟨2*k+1, Nat.odd_iff.mpr..., le⟩ : {m // Odd m ∧ 3 ≤ m}). Proofs: Odd (2*k+1) ✓ `Nat.odd_iff`/`odd_two_mul_add_one`? `Nat.odd_iff : Odd n ↔ n % 2 = 1`. There's `Nat.Odd`? For ℕ, `Odd (2*k+1)` : `odd_iff_exists_bit1`? Simply `⟨k, rfl⟩ : Odd (2*k+1)`? Odd n is ∃ m, n = 2*m+1 — so `⟨k, rfl⟩` ✓. And 3 ≤ 2*k+1 ⟸ 1 ≤ k ✓.\n\nsum_le_tsum s nonneg hsumm gives ∑ ≤ tsum.\n\nThen:\n∑ k ∈ Icc 1 K, g(2k+1) ≥ ∑ k ∈ Icc 1 K, (n/(2k+1) − 1)   [Finset.sum_le_sum with g_ge, 2k+1 ≥ 1 ✓]\n= n * harmonicOdd K − K.\n\nharmonicOdd K = H(2K+1) − H(K)/2 − 1.\n\nH(2K+1) ≥ log(2K+1) + γ (2K+1 ≥ 1 ✓), H(K) ≤ log(K+1) + γ (any K).\n\nSo harmonicOdd K ≥ log(2K+1) − log(K+1)/2 + γ/2 − 1.\n\nThen n * harmonicOdd K − K ≥ n*(log(2K+1) − log(K+1)/2 + γ/2 − 1) − K.\n\nSInfty n = (1−a1)*n − 1 + tsum ≥ (1−a1)*n − 1 + [that].\n\nFinal algebra: show\n(1−a1)*n − 1 + n*(log(2K+1) − log(K+1)/2 + γ/2 − 1) − K ≥ ½ n log n + (C1 − γ/2)*n − 4.\n\nRearrange: n*(log(2K+1) − log(K+1)/2) + (γ/2 − 1)*n − K − 1 + (1−a1)*n − (C1 − γ/2)*n − ½ n log n + 4 ≥ 0.\n\n(1−a1) − (C1 − γ/2) = γ/2 − 3/2 + ½ log 2 (identity to be verified by simp/ring with log_div, log_mul for log 4π = log 4 + log π, log 4 = 2 log 2, log 2π = log 2 + log π).\n\nSo the expression becomes:\nn*(log(2K+1) − log(K+1)/2 − ½ log n − ½ log 2) + (γ − 2)*n/... hmm wait let me redo: (γ/2 − 1)n + (γ/2 − 3/2 + ½log2)n = (γ − 5/2 + ½ log 2)n? No wait — I need to recompute. Let me redo the algebra cleanly.\n\nTarget: ½n log n + (C1−γ/2)n − 4 ≤ (1−a1)n − 1 + T where T ≥ n·A − K, A ≥ log(2K+1) − ½log(K+1) + γ/2 − 1.\n\nSuffices: ½n log n + (C1−γ/2)n − 4 ≤ (1−a1)n − 1 + n(log(2K+1) − ½log(K+1) + γ/2 − 1) − K.\n\nMove LHS: 0 ≤ (1−a1)n − (C1−γ/2)n + n(log(2K+1) − ½log(K+1)) + (γ/2 −1)n − K − ½n log n + 3.\n\n(1−a1) − (C1−γ/2) = γ/2 − 3/2 + ½log2 (identity).\n\n0 ≤ (γ/2 − 3/2 + ½log2)n + (γ/2 − 1)n + n(log(2K+1) − ½log(K+1) − ½log n − ½ log 2) + ½n log n... \n\nwait I made an error: −½n log n stays. Let me group:\n\n0 ≤ [(γ/2−3/2+½log2) + (γ/2−1)]n + n·log(2K+1) − (n/2)log(K+1) − (n/2)log n + ... \n\nHmm, I previously had: sufficies n·A ≥ ½n log n + (γ/2 − 3/2 + ½log2)n + K − 3. With A ≥ log(2K+1) − ½log(K+1) + γ/2 − 1:\n\nn(log(2K+1) − ½log(K+1) + γ/2 − 1) − (γ/2 − 3/2 + ½log2)n ≥ n(log(2K+1) − ½log(K+1) − ½log2 − ½) hmm this doesn't match my earlier derivation. Let me redo carefully.\n\nn(γ/2 − 1) − (γ/2 − 3/2 + ½log2)n = n[(γ/2 − 1) − γ/2 + 3/2 − ½log2] = n(½ − ½ log 2).\n\nSo suffices: n(log(2K+1) − ½log(K+1)) + n(½ − ½log2) ≥ ½n log n + K − 3\n⟺ n(log(2K+1) − ½log(K+1) − ½log n − ½log2 + ½) ≥ K − 3. \n\nHmm, earlier I wrote the bracket without the +½... earlier I had: \"⟺ n(log(2K+1) − ½ log(K+1) − ½ log n − ½ log 2) ≥ −n/2 + K − 3\" — yes same thing (moving ½n to the right). OK consistent.\n\nEven case (n = 2q, K = q−1): bracket = log(n−1) − ½log(n/2) − ½log n − ½log2 + ½ = log(n−1) − ½(log n − log 2) − ½log n − ½log2 + ½ = log(n−1) − log n + ½.\nSo need: n(log(n−1) − log n + ½) ≥ K − 3 = q − 4 = n/2 − 4.\nn(log(n−1) − log n) ≥ −n/(n−1). So LHS ≥ −n/(n−1) + n/2. Need −n/(n−1) + n/2 ≥ n/2 − 4 ⟺ 4 ≥ n/(n−1) ⟺ n ≥ 4/3 ✓.\n\nOdd case (n = 2q+1, K = q): bracket = log n − ½log((n+1)/2) − ½log n − ½log2 + ½ = ½log n − ½log(n+1) + ½.\nNeed: n(½log n − ½log(n+1) + ½) ≥ K − 3 = q − 3 = (n−1)/2 − 3 = n/2 − 7/2.\nn·½(log n − log(n+1)) ≥ −½. LHS ≥ n/2 − ½ ≥ n/2 − 7/2 ✓.\n\nGood, clean. In Lean, the final steps are field_simp/ring_nf + the log lemmas. The γ cancellation must be handled by ring (γ is just a real variable). The log 2 / log 4π / log 2π manipulations: need log(4π) = log 4 + log π = 2 log 2 + log π (Real.log_mul with positivity; log 4 = log(2²) = 2 log 2 via Real.log_pow or log_sq), log(2π) = log 2 + log π.\n\nThis is all doable. The proof will be long (~300 lines). Strategy: write helper lemmas as `private lemma`s in LiSkeleton.lean before the theorem:\n\n1. `harmonicOdd` def + `harmonicOdd_eq` (induction identity).\n2. `harmonic_lower` (H1), `harmonic_upper` (H2).\n3. `sInfTerm` bounds (g_nonneg, g_le_C, g_ge).\n4. `sInf_summable`.\n5. Main theorem.\n\nCasting harmonic: Mathlib `harmonic : ℕ → ℚ`. eulerMascheroniSeq uses `(harmonic n : ℝ)` via coercion ℝ of ℚ. In my identities I need: ((harmonic (2K+1) : ℚ) : ℝ) etc. harmonic_succ : harmonic (n+1) = harmonic n + (n+1)⁻¹ in ℚ. push_cast should turn ((harmonic n : ℚ):ℝ) identities into real ones: Rat.cast_add, Rat.cast_inv, Rat.cast_natCast.\n\nPotential pitfall: `harmonic n` coerced — simp lemmas `harmonic_zero`, `harmonic_one`, `harmonic_succ` are in ℚ; when goal is over ℝ with ↑, use `push_cast [harmonic_succ]`? Or `norm_cast`/`exact_mod_cast`. I'll manage at write time.\n\nNote the file has `noncomputable section` and `namespace NguyenRH1`, with `open Zeta23 Complex Filter Topology`.\n\nSInfty def uses `tsum fun m : {m : ℕ // Odd m ∧ 3 ≤ m} => ((1 - 1/(m:ℝ))^n - 1 + (n:ℝ)/(m:ℝ))` — with (m : ℝ) cast of m.val. My g defined as fun m : ℕ => (1 − 1/(m:ℝ))^n − 1 + n/(m:ℝ), then the tsum family is fun m : subtype => g m.val. ✓ (need to unfold SInfty).\n\nOne more check: the theorem statement has `(1/2) * n * Real.log n` where n : ℕ cast to ℝ, log n = log(↑n). ✓\n\nAlso for sum_le_tsum I need signature check: `Summable.sum_le_tsum (s : Finset ι) (hf : ∀ i ∉ s, 0 ≤ f i) (hf2 : Summable f)`? From lpSpace usage: `Summable.sum_le_tsum _ (fun i _ => this i) ?_` — hmm, there it's called on a Summable instance? `refine Summable.sum_le_tsum _ (fun i _ => this i) ?_; exact (lp.memℓp f).summable hp` — so the last explicit arg is the Summable proof. Signature: `theorem Summable.sum_le_tsum {f : ι → α} (s : Finset ι) (hf : ∀ i, i ∉ s → 0 ≤ f i) (hf₂ : Summable f) : ∑ i ∈ s, f i ≤ ∑' i, f i`. ✓\n\nFinset.sum_image signature: `Finset.sum_image (h : ∀ x ∈ s, ∀ y ∈ s, f x = f y → x = y) : ∑ x ∈ s.image f, g x = ∑ x ∈ s, g (f x)`. ✓\n\nNow, time to write. I'll write the code incrementally and build. Given the length, maybe write the helpers first, build, then the main theorem.\n\nAlso — check `Finset.sum_Icc_succ_top` name. And `Real.log_le_sub_one_of_pos`. And `Nat.even_or_odd`. And Even/Odd dest constructors: `Even n : ∃ r, n = r + r`; obtain ⟨q, rfl⟩ gives n = q + q; then `two_mul q : 2*q = q+q` to rewrite. Odd: `Odd n : ∃ k, n = 2*k+1`.\n\nFor q+q ≥ 20 → q ≥ 10: omega after linarith? omega handles nat arithmetic ✓.\n\nReal.log_div : `Real.log_div (hx : x ≠ 0) (hy : y ≠ 0) : log (x/y) = log x − log y` ✓.\n\nlog(n/(n−1)) ≤ 1/(n−1): Real.log_le_sub_one_of_pos : 0 < x → Real.log x ≤ x − 1. With x = n/(n−1): n/(n−1) − 1 = 1/(n−1) via field_simp.\n\nLet me also double check `one_add_mul_le_pow` is for linear ordered ring R with a : R — in Mathlib.Algebra.Order.Ring.Pow, variables likely {R} [LinearOrderedRing R]. ℝ qualifies ✓. Statement: `one_add_mul_le_pow (H : -2 ≤ a) (n : ℕ) : 1 + n * a ≤ (1 + a) ^ n`. With a = -1/m: 1 + n*(-(1/m)) ≤ (1 + -(1/m))^n. Need to massage: 1 + n*(-(1/m)) = 1 − n/m by ring; (1 + -(1/m)) = (1 − 1/m) by ring. So `by have := one_add_mul_le_pow (show (-2:ℝ) ≤ -(1/m) by ...) n; ...` — wait, is `a` explicit? `lemma one_add_mul_le_pow (H : -2 ≤ a) (n : ℕ)` — a implicit ✓.\n\n−2 ≤ −(1/m): since 1/m ≤ 1 ≤ 2 for m ≥ 1: (1:ℝ)/m ≤ 1 ⟺ 1 ≤ m ✓ `div_le_one (by positivity) |>.mpr`... `one_div_le_one_of_le`? just field_simp/positivity + nlinarith. Fine.\n\nAlso for term_upper: 1 − y^j ≤ j/m: from y^j ≥ 1 + j*(−1/m) (Bernoulli, needs −2 ≤ −1/m ✓) ⟹ 1 − y^j ≤ j/m ✓.\n\ngeom: 1 − y^n = (1−y) * Σ_{j ∈ range n} y^j. From geom_sum_mul y n: (Σ y^j)*(y−1) = y^n − 1. So (Σ y^j)*(1−y) = 1 − y^n: `neg_mul_neg`... (y−1) = −(1−y); (Σ)*(y−1) = −(Σ)*(1−y) = y^n−1 ⟹ (Σ)*(1−y) = 1−y^n. linarith/ring manipulation: have h := geom_sum_mul y n; then 1 − y^n = (1−y) * Σ by linear_combination -h? `linear_combination -h` should work, or ring_nf at h ⊢. I'll handle at write time.\n\nThen term = y^n − 1 + n/m = n/m − (1−y^n) = n/m − (1/m)·Σ y^j = (1/m)(n − Σ y^j) = (1/m) Σ_{j<n} (1 − y^j). Since 1−y = 1/m (field: y = 1 − 1/m, so 1 − y = 1/m by ring).\n\nSteps: term ≤ (1/m)·Σ_{j<n} (j/m) = (1/m)·(1/m)·Σ j = (1/m²)·(n(n−1)/2). Using:\n- `Finset.sum_congr rfl` with 1 − y^j ≤ j*(1/m).\n- Σ_{j ∈ range n} (j:ℝ)*(1/m) = (1/m)*Σ j: `Finset.sum_mul`? (∑ j) * (1/m): `Finset.sum_mul : (∑ i ∈ s, f i) * b = ∑ i ∈ s, f i * b` ✓.\n- Σ_{j ∈ range n} (j:ℝ) = n(n−1)/2: from `Finset.sum_range_id_mul_two n : (∑ i ∈ range n, i) * 2 = n * (n − 1)` (ℕ), cast to ℝ: push_cast, then linarith/ring to get the real sum = n(n−1)/2.\n\nAlso nonneg: each 1 − y^j ≥ 0: y^j ≤ 1: `pow_le_one₀ (hy0 : 0 ≤ y) (hy1 : y ≤ 1)`. y = 1 − 1/m: 0 ≤ y ⟺ 1/m ≤ 1 ✓ m ≥ 1; y ≤ 1 ⟺ 0 ≤ 1/m ✓.\n\nOK. One concern: `Summable.of_nonneg_of_le` exact signature: `Summable.of_nonneg_of_le (hg : ∀ b, 0 ≤ g b) (h : ∀ b, g b ≤ f b) (hf : Summable f) : Summable g` ✓ I believe.\n\nThe p-series: `summable_one_div_nat_pow {p : ℕ} : Summable (fun n : ℕ => 1 / (n : ℝ)^p) ↔ 1 < p`. Hmm is the family `fun n => 1/(n:ℝ)^p`? At n=0: 1/0^2 = 1/0 = 0 ✓ fine. Let me double-check the exact form by reading PSeries.lean:330.\n\nAlso `tsum` in SInfty is over ℝ so it's `∑'`. `Summable.sum_le_tsum` gives ∑ s ≤ ∑' ✓.\n\nLet me check PSeries and a few lemma names by reading the file, then write the Lean code.","signature":""},{"type":"text","text":"All the pieces exist. Let me verify the exact statements of a few more, then write the proof."},{"type":"tool_use","id":"Read_28","name":"Read","input":{"file_path":"D:\\Python\\Quantum\\zeta-23-lean\\.lake\\packages\\mathlib\\Mathlib\\Analysis\\PSeries.lean","offset":320,"limit":25},"extra_content":null},{"type":"tool_use","id":"Grep_29","name":"Grep","input":{"pattern":"theorem Summable.of_nonneg_of_le|theorem Summable.sum_le_tsum|theorem sum_Icc_succ_top|theorem sum_range_id_mul_two|theorem log_le_sub_one_of_pos","path":"D:\\Python\\Quantum\\zeta-23-lean\\.lake\\packages\\mathlib\\Mathlib","output_mode":"content","-n":true,"-A":2,"head_limit":40},"extra_content":null}]