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EXERCISES - LIMIT OF SEQUENCES

limn→∞ (n+4)/n = 1; ∀ ε > 0, ∃ n0 = n0(ε) | ∀ n ≥ n0 ⇒ |an-l| < ε; ∀ ε > 0, ∃ n0 = n0(ε) | ∀ n ≥ n0 ⇒ |((n+4)/n)-1| < ε; |(n+4-n)/n| < ε; |4/n| < ε; 4/n < ε; n > 4/ε; n0 > [4/ε]+1

limn→∞ (n3/2+1)/(2n-1); n3/2+1 ~ n3/2, 2n-1 ~ 2n; limn→∞ (n3/2+1)/(2n-1) = limn→∞ n3/2/2n = limn→∞ n1/2/2 = limn→∞n/2 = +∞; √n/2 is a harmonic sequence with α = 1/2; {nα}n∈ℕ, α ∈ ℝ; limn→∞ nα = {+∞, α > 0; 1, α = 0; 0, α < 0}

limn→∞ (n4+3n-1/n+2)/(n2+1/n2) = limn→∞ n4/n2 = limn→∞ n2 = +∞

limn→∞ (1+√n)/(n+1)2nα, α ∈ ℝ; 1+√n ~ n1/2; (n+1)2 ~ n2; nα ~ nα; limn→∞ (1+√n)/(n+1)2nα = limn→∞ n1/2/n2nα = limn→∞ n1/2-2-α = limn→∞ n-3/2-α = {+∞, -3/2-α > 0 ⇔ α < -3/2; 1, -3/2-α = 0 ⇔ α = -3/2; 0, -3/2-α < 0 ⇔ α > -3/2}

Geometric sequence: limn→∞ qn = {+∞, q > 1; 1, q = 1; 0, -1 < q < 1; ∄, q ≤ -1}

Harmonic sequence: limn→∞ nα = {+∞, α > 0; 1, α = 0; 0, α < 0}

limn→∞ 3n+4n-5n = +∞-∞ = limn→∞ 5n(3n/5n+4n/5n-5n/5n) = limn→∞ 5n((3/5)n+(4/5)n-1) = -∞; 5n→∞, (3/5)n→0, (4/5)n→0

limn→∞ (2n+1+1)/(3n+n) = limn→∞ 2n+1/3n = limn→∞ 2n2/3n = limn→∞ 2(2/3)n = 0

limn→∞ (2n-4n)/(3n-n!) = limn→∞ 4n(2n/4n-4n/4n)/n!(3n/n!-n!/n!) = limn→∞ 4n((1/2)n-1)/n!(3n/n!-1) = 0; (1/2)n-1→-1, 3n/n!-1→-1

limn→∞ ((n2+n)/(n2-n+2))n = 1 [1] = limn→∞ (1+(n2+n)/(n2-n+2)-1)n = limn→∞ (1+(n2+n-n2+n-2)/(n2-n+2))n = limn→∞ (1+1/(n2-n+2)/(2n-2))n [2] = limn→∞ ((1+1/(n2-n+2)/(2n-2))(n2-n+2)/(2n-2))((2n-2)/(n2-n+2))n [3] = e2; [1] limn→∞ (1+x/an)an = ex, limn→∞ an = +∞; [2] limn→∞ (n2-n+2)/(2n-2) = +∞; [3] limn→∞ (1+1/(n2-n+2)/(2n-2))(n2-n+2)/(2n-2) = e, limn→∞ = n(2n-2)/(n2-n+2) = limn→∞ 2n2/n2 = 2

limn→∞ n⋅tan(1/n) = ∞⋅0 = limn→∞ n⋅sin(1/n)/cos(1/n) = limn→∞ (sin(1/n)/1/n)(1/cos(1/n)) = 1; limn→∞ sin(an)/an = 1, limn→∞ an = 0

limn→∞ (1-cos(3/n))/sin(3/n2) = 0/0 = limn→∞ ((1-cos(3/n))/(3/n)2)((3/n)23/n2/sin(3/n2)3/n2) = limn→∞ ((1-cos(3/n))/(3/n)2)(3/n2/sin(3/n2))(9/n2)(n2/3) = 3/2; limn→∞ (1-cos(an))/an2 = 1/2, limn→∞ an = 0; limn→∞ sin(an)/an = 1, limn→∞ an = 0

limn→∞ (21/n-1)/1/n = limn→∞ ((e(1/n)ln(2)-1)/(1/n)ln(2))ln(2) = ln(2); ab = eb⋅ln(a); limn→∞ (ean-1)/an = 1, limn→∞ an = 0

limn→∞ (2n-1)/n = limn→∞ 2n(1-1/2n)/n = +∞; A geometric sequence diverges to +∞ more rapidly than a harmonic sequence

limn→∞ (1+1/nn)n! = 1 = limn→∞ ((1+1/nn)nn)n!/nn = e0 = 1; limn→∞ (1+x/an)an = ex, limn→∞ an = +∞; (ab)c = abc

limn→∞n+2-√n+1 = +∞-∞ = limn→∞ (√n+2-√n+1)(√n+2+√n+1)/(√n+2+√n+1) = limn→∞ ((n+2)-(n+1))/√n(√1+2/n+√1+1/n) = 1/2√n = 1/∞ = 0; (a-b)(a+b) = a2-b2

limn→∞ nn

Applying the Cesàro criterion: limn→∞ nan, an = n

limn→∞ an+1/an = limn→∞ (n+1)/n = 1 ⇒ limn→∞ nn = 1

limn→∞ nn!/(2n+1); applying the Cesàro criterion: an = n!/(2n+1); limn→∞ an+1/an = limn→∞ (n+1)!/(2n+1+1)(2n+1)/n! [1] = limn→∞ (n!(n+1)/(2n+1+1))((2n+1)/n!) = limn→∞ ((n+1)/(2n+1+1))(2n+1) = limn→∞ (n+1)(2n(1+1/2n)/2n+1(1+1/2n+1)) = limn→∞ (n+1)(2n(1+1/2n)/2⋅2n(1+1/2n+1)) = limn→∞ (n+1)/2 = +∞; limn→∞ nn!/(2n+1) = +∞; [1] (n+1)! = n!(n+1)

limn→∞ nC(2n, n)); binomial coefficient: n, k ∈ ℕ | n ≥ k ≥ 0; C(n, k) = n!/k!(n-k)!; applying the Cesàro criterion: an = C(2n, n); limn→∞ an+1/an = limn→∞ C(2(n+1), n+1)/C(2n, n) = limn→∞ C(2n+2, n+1)/C(2n, n) = limn→∞ ((2n+2)!/(n+1)!(2n+2-n-1)!)/((2n)!/n!(2n-n)!) = limn→∞ ((2n+2)!/(n+1)!(n+1)!)(n!n!/(2n)!) [1][2] = limn→∞ ((2n)!(2n+1)(2n+2)/n!(n+1)n!(n+1))(n!n!/(2n)!) = limn→∞ (2n+1)(2n+2)/(n+1)(n+1) = limn→∞ (2n+1)(2n+2)/(n+1)2 = limn→∞ (2n)2(1+1/2n)(1+2/2n)/n2(1+1/n)2 = limn→∞ 4n2/n2 = 4; limn→∞ nC(2n, n) = 4; [1] (n+1)! = n!(n+1); [2] (2n+2)! = (2n)!(2n+1)(2n+2)

limn→∞ n2n!; 1 = n21n2n!n2nn = (nn)1/n2 = n1/n; limn→∞ 1 ≤ limn→∞ n2n! ≤ limn→∞ n1/n; 1 ≤ limn→∞ n2n! ≤ 1; limn→∞ n2n! = 1

{an}n∈ℕ | {|an|}n∈ℕ is decreasing, then: a) {an}n∈ℕ is monotone, b) limn→∞ an ∃, c) limn→∞ an ∄, d) {an}n∈ℕ is bounded; a) {an}n∈ℕ is monotone, false, considering {an}n∈ℕ = {(-1)n}, {|(-1)n|} = {1}n∈ℕ, a constant sequence is both increasing and decreasing, {|(-1)n|} is decreasing, but (-1)n is not a monotone sequence; b) limn→∞ an ∃, false, considering {an}n∈ℕ = {(-1)n}, {|(-1)n|} is decreasing, but limn→∞ (-1)n ∄; c) limn→∞ an ∄, false, considering {an}n∈ℕ = {1/n} | {|1/n|}n∈ℕ = {1/n} is decreasing and limn→∞ 1/n = 0; d) {an}n∈ℕ is bounded, true, {|an|}n∈ℕ is decreasing ⇒ |a0| ≥ |a1| ≥ ... ≥ |an| ∀ n ∈ ℕ ⇔ |an| ≤ |a0| ∀ n ∈ ℕ ⇔ -|a0| ≤ an ≤ |a0| ∀ n ∈ ℕ, {an}n∈ℕ is bounded if ∃ m, M ∈ ℝ | m ≤ an ≤ M ∀ n ∈ ℕ

{an} = (2n+3)(-1)n: a) {an} converges, b) {an} diverges, c) {an} is bounded but does not converge, d) {an} is not bounded, but does not diverge; (2n+3)(-1)n = {(2n+3), n = 2k; 1/(2n+3), n = 2k+1}; the sequence diverges to +∞ with n even, converges to zero with n odd; the sequence is not bounded and it is irregular, so is true d) {an} is not bounded, but does not diverge

{an}n∈ℕ is a bounded sequence: a) |an| ≤ 1000 ∀ n ∈ ℕ, b) {an}n∈ℕ is not increasing, c) ∃ α > 0 | 2-an < α ∀ n, d) limn→∞ an ∃ as a finite number, the sequence converges; a) |an| ≤ 1000 ∀ n ∈ ℕ, false, an = 1001 ∀ n is bounded, but it does not verify a); b) {an}n∈ℕ is not increasing, false, an = -1/n is bounded, -1 ≤ -1/n ≤ 0, and it is increasing; d) limn→∞ an ∃ as a finite number, the sequence converges, false, {sin(n)}n∈ℕ is bounded, but it is irregular, in fact if ∃ limn→∞ an ⇒ {an} is bounded, but if {an} is bounded !⇒ ∃ limn→∞ an; c) ∃ α > 0 | 2-an < α ∀ n, true

a0 = 1, an+1 = √1+an: a) verify by induction that {an} is increasing, b) verify by induction that 1 ≤ an ≤ 2 ∀ n ∈ ℕ, c) calculate limn→∞ an; a) verify by induction that {an} is increasing; thesis: an ≤ an+1 ∀ n ∈ ℕ, basis of induction (n = 0): a0 ≤ a1, a0 = 1, a1 = √1+a0 = √1+1 = √2, 1 ≤ √2; inductive step: hypothesis an ≤ an+1, thesis an+1 ≤ an+2, an+2 = √1+an+1 ≥ √1+an = an+1, an+2 ≥ an+1; b) verify by induction that 1 ≤ an ≤ 2 ∀ n ∈ ℕ; considering an ≥ 1 ∀ n ∈ ℕ; basis of induction (n = 0): a0 ≥ 1 ⇔ 1 ≥ 1; inductive step: hypothesis an ≥ 1, thesis an+1 ≥ 1, an+1 = √1+an ≥ √1+1 = √2 ≥ 1; c) calculate limn→∞ an; a) + b) ⇒ ∃ limn→∞ an = l ⇒ {limn→∞ an+1 = l; limn→∞ an+1 = limn→∞1+an = limn→∞1+l} ⇔ l = √1+l ⇔ l2 = 1+l ⇔ l2-l-1 = 0, using the quadratic formula ax2+bx+c = 0, x = (-b±√b2-4ac)/2a, l2-l-1 = 0, (1±√1+4)/2, (1±√5)/2, 1 ≤ an ≤ 2 ⇒ 1 ≤ l ≤ 2 for the sign permanence theorem, but (1-√5)/2 < 0 ⇒ l = (1+√5)/2 that is the golden ratio