(26) 2/(4+3) is (A) convergent; (B) divergent; (C) None of these. (27) (n+2)/(n+1) is (A)...

 

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(26) 2/(4+3¹) is (A) convergent; (B) divergent; (C) None of these. (27) Σ (n²+2)/(n³+1) is (A) convergent; (B) divergent; (C) None of these. (28) Σ 1/3 n(n+1)(n+2) is (A) convergent; (B) divergent; (C) None of these. (29) Let an = n/(n4 + 1)1/2 and b₁ = 1/n, then lim (an/bn) (A) 1; (B) ∞o; (C) 1/2; (D) 0; (E) None of these. (30) Let an = n/(4n6+1)1/2 and b₁ = 1/n², then lim (a/b) = (A) 1; (B) ∞o; (C) 1/2; (D) 0; (E) None of these. (31) Σ sin(1/n²) is (A) convergent; (B) divergent; (C) None of these. = (32) lim n¹/n = lim e[(Inn)/n] = elim[(In x)/x] = (A) 0; (B) ∞o; (C) 1; (D) e; (E) None of these. (26) 2/(4+3¹) is (A) convergent; (B) divergent; (C) None of these. (27) Σ (n²+2)/(n³+1) is (A) convergent; (B) divergent; (C) None of these. (28) Σ 1/3 n(n+1)(n+2) is (A) convergent; (B) divergent; (C) None of these. (29) Let an = n/(n4 + 1)1/2 and b₁ = 1/n, then lim (an/bn) (A) 1; (B) ∞o; (C) 1/2; (D) 0; (E) None of these. (30) Let an = n/(4n6+1)1/2 and b₁ = 1/n², then lim (a/b) = (A) 1; (B) ∞o; (C) 1/2; (D) 0; (E) None of these. (31) Σ sin(1/n²) is (A) convergent; (B) divergent; (C) None of these. = (32) lim n¹/n = lim e[(Inn)/n] = elim[(In x)/x] = (A) 0; (B) ∞o; (C) 1; (D) e; (E) None of these.

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A convergent 27 A convergent 28 A convergent 29 A 1 30 D 0 31 A convergent 32 A 0 why 26 243 is conv... View the full answer