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LIMIT OF A SEQUENCE


{an}n∈ℕ, l ∈ ℝ, the sequence converges to l as n tends to +∞, written as limn→+∞ an = l, if ∀ ε > 0, ∃ n0 = n0(ε) ∈ ℕ | ∀ n ≥ n0 then |an-l| < ε

|x| < r ⇔ -r < x < r; |an-l| < ε ⇔ -ε < an-l < ε ⇔ l-ε < an < l+ε; ∀ ε > 0, |an-l| < ε definitely

limn→+∞ 1/n = 0 ⇔ ∀ ε > 0, ∃ n0 | ∀ n > n0, |(1/n)-0| < ε; |1/n| < ε ⇔ -ε < 1/n < ε ⇔ 1/n < ε ⇔ n > 1/ε; n0 = [1/ε] + 1 that is the first natural number greater than 1/ε; [x] = n, if n ≤ x < n+1, [x] = integer part of x

If limn→∞ an = 0 then {an}n∈ℕ is an infinitesimal sequence.

Theorem of uniqueness of the limit of a sequence: if the limit of a sequence exists, then it is unique.

Proof by contradiction of the theorem of uniqueness of the limit of a sequence: {an}n∈ℕ, l1, l2 ∈ ℝ, l1 ≠ l2 | limn→∞ an = l1 and limn→∞ an = l2; l1 ≠ l2 ⇒ |l1-l2| ≠ 0; ε := |l1-l2| > 0; ∃ n1 | ∀ n > n1, |an-l1| < ε/4; ∃ n2 | ∀ n > n2, |an-l2| < ε/4; n0 = max{n1,n2}; ∀ n > n0 = max{n1,n2}, ε = |l1-l2| = |l1-an+an-l2| [1] ≤ |l1-an|+|an-l2| = |an-l1|+|an-l2| < (ε/4)+(ε/4) = ε/2 ⇒ ε < ε/2 ⇒ 1 < 1/2 that is a contradiction; [1] Triangle inequality, |x+y| ≤ |x|+|y|

limn→∞ 1/n = 0; {1/n}n∈ℕ = {1, 1/2, 1/3, 1/4, ...}

limn→∞ n2; {n2}n∈ℕ = {0, 1, 4, 9, ...}

{an}n∈ℕ diverges to +∞ for n→+∞, limn→+∞ an = +∞, if ∀ M ∈ ℝ, ∃ n0 = n0(M) ∈ ℕ | ∀ n ≥ n0 then an > M

{an}n∈ℕ diverges to -∞ for n→+∞, limn→+∞ an = -∞, if ∀ M ∈ ℝ, ∃ n0 = n0(M) ∈ ℕ | ∀ n ≥ n0 then an < M

{an}n∈ℕ is divergent if it diverges to ±∞

limn→+∞ an = +∞ if ∀ M ∈ ℝ, an > M definitely

limn→+∞ an = -∞ if ∀ M ∈ ℝ, an < M definitely

limn→∞ n2 = +∞; ∀ M ∈ ℝ, ∃ n0 = n0(M) ∈ ℕ | ∀ n ≥ n0, n2 > M; n2 > M ⇔ {∀ n ∈ ℕ, M < 0; n > √M, M ≥ 0}; n > √M ⇔ n ≥ [√M]+1 = n0(M)

limn→∞ -n3 = -∞

{(-1)n}n∈ℕ

If {an}n∈ℕ neither converges to l ∈ ℝ, nor diverges, then it is irregular or oscillating.

{an} is regular if it is convergent or divergent, or it is irregular if it is oscillating.

(-1)n⋅n = {n, n = 2k; -n, n = 2k+1}; this sequence is irregular or oscillating {0, -1, 2, -3, 4, -5, 6, -7, 8, -9, ...}

(-n)n = (-1)n⋅nn; this sequence is irregular or oscillating {0, -1, 4, -27, 256, -3125, ...}

an = {1, n = 2k; 1/n, n = 2k+1}; this sequence is irregular or oscillating {1, 1, 1, 1/3, 1, 1/5, ...}

{sin(n)}n∈ℕ is an irregular sequence, because for n that tends to infinity it does not converge or diverge; ∄ limn→∞ sin(n)

{cos(n)}n∈ℕ is an irregular sequence, because for n that tends to infinity it does not converge or diverge; ∄ limn→∞ cos(n)

limn→∞ (nn+n!)/(en+2n) = +∞

Geometric sequence {qn}n∈ℕ, q ∈ ℝ; limn→∞ qn = {+∞, q > 1; 1, q = 1; 0, -1 < q < 1; ∄, q ≤ -1}; limn→∞ 1n = limn→∞ 1 = 1; q > 1 ⇒ q = 1+(q-1), q-1 > 0, qn = (1+(q-1))n [1] ≥ 1+n(q-1) ∀ n ∈ ℕ, [1] Bernoulli's inequality (1+x)n ≥ 1+nx ∀ x > -1, limn→∞ 1+n(q-1) = +∞ ⇒ limn→∞ qn = +∞; q < -1 ⇒ -q > 1, qn = (-1(-q))n = (-1)n(-q)n, that is +∞ for n = 2k and -∞ n = 2k+1

Harmonic sequence {nα}n∈ℕ, α ∈ ℝ; limn→∞ nα = {+∞, α > 0; 1, α = 0; 0, α < 0}; limn→∞ n3/2 = +∞; limn→∞ 1/n2 = 0, 1/n2 = n-2; limn→∞ n0 = limn→∞ 1 = 1

If {an}n∈ℕ is convergent, then it is bounded, that is ∃ m, M ∈ ℝ | m ≤ an ≤ M ∀ n ∈ ℕ

{(-1)n} is bounded, -1 ≤ (-1)n ≤ 1 ∀ n ∈ ℕ, but ∄ limn→∞ (-1)n

How can we calculate the limit of a sequence? To apply the definition of limit limn→∞ an = l, we need to know l ∈ ℝ.

The limit of a sum is the sum of the limits: limn→∞ (an+bn) = l+m; limn→∞ an = l, limn→∞ bn = m, l,m ∈ ℝ

The limit of a difference is the difference of the limits: limn→∞ (an-bn) = l-m; limn→∞ an = l, limn→∞ bn = m, l,m ∈ ℝ

The limit of a product is the product of the limits: limn→∞ (an⋅bn) = l⋅m; limn→∞ an = l, limn→∞ bn = m, l,m ∈ ℝ

The limit of a ratio is the ratio of the limits: limn→∞ (an/bn) = l/m, m ≠ 0; limn→∞ an = l, limn→∞ bn = m, l,m ∈ ℝ

The limit of a power is the power of the limits: limn→∞ (an)^bn = lm, l > 0; limn→∞ an = l, limn→∞ bn = m, l,m ∈ ℝ

The limit of a module is the module of the limit: limn→∞ |an| = |l|; limn→∞ an = l, l ∈ ℝ

limn→∞ (n2-n)/(n2-1) = limn→∞ n2(1-(1/n))/n2(1-(1/n2)) = limn→∞ (1-(1/n))/(1-(1/n2)) = 1/1 = 1; limn→∞ 1/n = limn→∞ n-1 = 0; limn→∞ 1/n2 = limn→∞ n-2 = 0

limn→∞ an = l ∈ ℝ, limn→∞ bn = ±∞, limn→∞ an+bn = ±∞; l±∞ = ±∞

Extended set of real numbers: = ℝ ∪ {±∞}

l±∞ = ±∞ ∀ l ∈ ℝ

+∞+∞ = +∞; limn→∞ an = +∞, limn→∞ bn = +∞ ⇒ limn→∞ an+bn = +∞

-∞-∞ = -∞; limn→∞ an = -∞, limn→∞ bn = -∞ ⇒ limn→∞ an+bn = -∞

+∞⋅l = {+∞, l > 0; -∞, l < 0}

l/±∞ = 0 ∀ l ∈ ℝ

q+∞ = {+∞, q > 1; 0, q < 1}

+∞⋅+∞ = +∞

-∞⋅-∞ = +∞

+∞⋅-∞ = -∞

-∞⋅+∞ = -∞

limn→∞ (n2)+(1/n)+(3n) = +∞+0+∞ = +∞

limn→∞ (n2+2n-1)/(n5+3n+2) = limn→∞ (n2(1+(2/n)-(1/n2)))/(n5(1+(3/n4)+(2/n5))) = (1+(2/n)-(1/n2))/(n3(1+(3/n4)+(2/n5))) = 1/+∞ = 0

Indeterminate Forms: +∞-∞; ±∞/±∞; 0/0; 0⋅(±∞); 1±∞; 00; ∞0; l/0, l ∈ ℝ

limn→∞ an = +∞, limn→∞ bn = -∞, limn→∞ an+bn = ?

an = n+1, bn = -n; limn→∞ an = +∞, limn→∞ bn = -∞ ⇒ limn→∞ an+bn = limn→∞ 1 = 1

an = n, bn = -n; limn→∞ an = +∞, limn→∞ bn = -∞ ⇒ limn→∞ an+bn = 0

An indeterminate form of a limit must be transformed into a determinate form: a limit is always either convergent, or divergent, or irregular.

an = n3, bn = n2; limn→∞ an/bn = limn→∞ n3/n2 = ∞/∞ = limn→∞ n = +∞; this sequence was initially considered indeterminate, but it turned out to be divergent.

an = 2n3, bn = n3; limn→∞ an/bn = limn→∞ 2n3/n3 = ∞/∞ = limn→∞ 2 = 2; this sequence was initially considered indeterminate, but it turned out to be convergent.

an = n, bn = n4; limn→∞ an/bn = limn→∞ n/n4 = ∞/∞ = limn→∞ 1/n3 = 1/∞ = 0; this sequence was initially considered indeterminate, but it turned out to be infinitesimal.

l/0, l ≠ 0, is an indeterminate form.

an = 1, bn = (-1)n/n; limn→∞ an = limn→∞ 1 = 1; limn→∞ bn = limn→∞ (-1)n/n = 0; limn→∞ an/bn = limn→∞ 1/(-1)n/n = 1/0 = limn→∞ n/(-1)n = limn→∞ (-1)nn = ∄; this sequence was initially considered indeterminate, but it turned out to be irregular.

0 is an indeterminate form; ab = eb·ln(a); ∞0 = e0·ln(∞) = e0·∞

If an is a bounded sequence, ∃ m,M ∈ ℝ | m ≤ an ≤ M ∀ n; limn→∞ bn = 0 ⇒ limn→∞ anbn = 0; a bounded sequence, multiplied by an infinitesimal sequence, is equal to zero.

If an is a bounded sequence, ∃ m,M ∈ ℝ | m ≤ an ≤ M ∀ n; limn→∞ bn = ±∞ ⇒ limn→∞ an/bn = 0; a bounded sequence, divided by a divergent sequence, is equal to zero.

limn→∞ sin(n)/n = 0; sin(n) is irregular and bounded because -1 ≤ sin(x) ≤ 1 ∀ x ∈ ℝ

limn→∞ cos(2n)/2n = 0; cos(2n) is irregular and bounded because -1 ≤ cos(x) ≤ 1 ∀ x ∈ ℝ ⇒ -1 ≤ cos(2n) ≤ 1 ∀ n ∈ ℕ; limn→∞ 1/2n = limn→∞ (1/2)n = 0, geometric sequence qn with -1 < q < 1

Comparison theorem: {an}n∈ℕ, {bn}n∈ℕ | {an} ≤ {bn} ∀ n ∈ ℕ, limn→∞ an = l, limn→∞ bn = m, l ≤ m

Squeeze theorem or sandwich theorem or theorem of carabinieri: {an}n∈ℕ, {bn}n∈ℕ, {cn}n∈ℕ | an ≤ cn ≤ bn ∀ n ∈ ℕ, limn→∞ an = limn→∞ bn = l, limn→∞ cn = l

Permanence of the sign: {an} | limn→∞ an = l; if an ≥ 0 ∀ n, then l ≥ 0; if an ∈ [α,β] ∀ n, then l ∈ [α,β]

If an > 0, then l > 0? No. For example {1/n}n∈ℕ, 1/n > 0, limn→∞ 1/n = 0

limn→∞ 1/(5n+cos(n)); 5n-1 ≤ 5n+cos(n) ≤ 5n+1; 1/(5n+1) ≤ 1/(5n+cos(n)) ≤ 1/(5n-1); 0 ≤ 1/(5n+cos(n)) ≤ 0; for the squeeze theorem limn→∞ 1/(5n+cos(n)) = 0

limn→∞ na = 1 ∀ a > 0; for a > 1, limn→∞ na = 1 ⇔ limn→∞ na-1 = 0; dn := na-1; we have to prove that limn→∞ dn = 0; dn = na-1 ⇔ na = 1+dn ⇔ a = (1+dn)n [1] ≥ 1+ndn ⇔ a ≥ 1+ndn ⇔ a-1 ≥ ndn ⇔ (a-1)/n ≥ dn ≥ 0 [2] ⇒ limn→∞ dn = 0; [1] Bernoulli's inequality, (1+x)n ≥ 1+nx ∀ x > -1, ∀ n ∈ ℕ; [2] for the squeeze theorem; for 0 < a < 1, limn→∞ na = limn→∞ 1/(1/na) = limn→∞ 1/n1/a) [1] = 1/1 = 1; [1] 0 < a < 1 ⇒ 1/a > 1, applying the previous case; the nth roots of the real numbers converge to 1.

{an}n∈ℕ | limn→∞ an = 0; if an ≥ 0 ∀ n ∈ ℕ, then limn→∞ an = 0+; if an ≤ 0 ∀ n ∈ ℕ, then limn→∞ an = 0-

l/0+ = {+∞, l > 0; -∞, l < 0}

l/0- = {+∞, l < 0; -∞, l > 0}

limn→∞ an = l, limn→∞ bn = 0+ ⇒ limn→∞ an/bn = {+∞, l > 0; -∞, l < 0}

limn→∞ 1/((1/2)n+1/n) = 1/0+ = +∞; (1/2)n+1/n ≥ 0

Monotone convergence theorem or theorem of regularity of monotone sequence: if {an}n∈ℕ is monotone, then it is regular, that is convergent or divergent; if {an}n∈ℕ is increasing, then limn→∞ an = sup{an | n ∈ ℕ}; if {an}n∈ℕ is decreasing, then limn→∞ an = inf{an | n ∈ ℕ}

A monotonous succession is always regular, it cannot be oscillating; if it is increasing converges to a real number or diverges to + ∞; if it is decreasing converges to a real number or diverges to -∞.

Demonstration of the monotone convergence theorem: suppose that l = sup{an | n ∈ ℕ} ∈ ℝ; thesis l = limn→∞ an ⇔ ∀ ε > 0, ∃ n0 | ∀ n > n0, |an-l| < ε ⇔ -ε < an-l < ε; by definition of supremum, l is a majorant or upper bound of {an | n ∈ ℕ}, therefore an ≤ l ∀ n ∈ ℕ ⇔ an-l ≤ 0 < ε ∀ n ∈ ℕ; by definition of supremum, l-ε is not a majorant or upper bound of {an | n ∈ ℕ} ⇒ ∃ n0 ∈ ℕ | l-ε < an0, because {an}n∈ℕ is increasing, an ≥ an0 ∀ n > n0, therefore -ε ≤ an-l ∀ n ≥ n0

{(1+1/n)n}n∈ℕ is the Euler's sequence.

The Euler's sequence {(1+1/n)n}n∈ℕ is bounded, 2 ≤ (1+1/n)n ≤ 3, and it is increasing.

limn→∞ (1+1/n)n := e, the Euler's number.

For the permanence of the sign 2 ≤ e ≤ 3

(1+1/n)n, increasing the value of n, there is a better approximation of the value of e.

e = 2.71828...

e ∈ ℝ \ ℚ, the Euler's number is irrational.

{an}n∈ℕ, {bn}n∈ℕ are asymptotic sequences if limn→∞ an/bn = 1; an ~ bn n→∞

an ~ bn ⇒ limn→∞ an = limn→∞ bn

an = n2-3n+2, bn = n2+1; limn→∞ an/bn = limn→∞ (n2-3n+2)/(n2+1) = limn→∞ n2(1-3/n+2/n2)/n2(1+1/n2) = limn→∞ (1-3/n+2/n2)/(1+1/n2) = 1/1 = 1 ⇒ n2-3n+2 ~ n2+1

an = n5/2+n2+n1/2+2, bn = n5/2; limn→∞ an/bn = limn→∞ (n5/2+n2+n1/2+2)/(n5/2) = limn→∞ 1+1/n1/2+1/n2+2/n5/2 = 1

limn→∞ an = limn→∞ bn !⇒ an ~ bn

an = n2, bn = n; limn→∞ an = limn→∞ bn = +∞; limn→∞ an/bn = limn→∞ n2/n = +∞

The asymptotic relation is an equivalence relation: it is reflexive, symmetrical, and transitive; reflexive: an ~ an ⇔ limn→∞ an/an = 1; symmetrical: an ~ bn ⇔ bn ~ an, limn→∞ bn/an = limn→∞ 1/an/bn = limn→∞ 1/1 = 1; transitive: an ~ bn, bn ~ cn ⇒ an ~ cn, limn→∞ an/cn = limn→∞ (an/bn)(bn/cn) = limn→∞ 1⋅1 = 1

Substitution principle: if an ~ an', bn ~ bn', cn ~ cn', then limn→∞ an⋅bn/cn = limn→∞ an'⋅bn'/cn'

limn→∞ (n3+2n+1)/(n4+4n2+3); n3+2n+1 ~ n3, n4+4n2+3 ~ n4; limn→∞ (n3+2n+1)/(n4+4n2+3) = limn→∞ n3/n4 = limn→∞ 1/n = 0

The substitution principle applies only to products and ratios: an ~ an', bn ~ bn' ⇒ limn→∞ an/bn = limn→∞ an'/bn', limn→∞ anbn = limn→∞ an'⋅bn'

The substitution principle does not generally apply to sums, differences and powers: an ~ an', bn ~ bn' !⇒ limn→∞ an±bn = limn→∞ an'±bn', limn→∞ an^bn = limn→∞ an'^bn'

an = n+1 ~ an' = n, bn = -n ~ bn' = -n; limn→∞ an+bn = limn→∞ n+1-n = 1; limn→∞ an'+bn' = limn→∞ n-n = 0

limn→∞ an = l ⇔ limn→∞ an/l = 1; an ~ l

an = 1+1/n ~ an' = 1, bn = n ~ bn' = n; limn→∞ an^bn = limn→∞ (1+1/n)n = e; limn→∞ an'^bn' = limn→∞ 1n = limn→∞ 1 = 1

limn→∞ ln(n)/nα = 0, α > 0; it would be ∞/∞, but the harmonic sequence at the denominator grows faster than the logarithm sequence at the numerator.

limn→∞ nα/qn = 0, α > 0, q > 1; it would be ∞/∞, but the geometric sequence at the denominator grows faster than the harmonic sequence at the numerator.

limn→∞ qn/n! = 0, q > 1; it would be ∞/∞, but the factorial sequence at the denominator grows faster than the geometric sequence at the numerator.

limn→∞ n!/nn = 0; it would be ∞/∞, but the nn sequence at the denominator grows faster than the factorial sequence at the numerator.

Infinity hierarchy from slowest to fastest: ln(n), nα, qn, n!, nn.

limn→∞ qn/nn = limn→∞ (qn/n!)(n!/nn) = 0

limn→∞ nn/qn = limn→∞ 1/qn/nn = 1/0+ = +∞

limn→∞ nn = 1; ab = eb⋅ln(a); limn→∞ nn = limn→∞ n1/n = limn→∞ e(ln(n))/n = limn→∞ e0 = 1

limn→∞ (1+x/an)an = ex, ∀ x ∈ ℝ, ∀ {an}n∈ℕ | limn→∞ an = +∞; limn→∞ (1+1/n)n = e, x = 1, an = n; limn→∞ (1+2/(n2+1))n2+1 = e2, x = 2, an = n2+1

limn→∞ sin(an)/an = 1, limn→∞ an = 0

limn→∞ (ean-1)/an = 1, limn→∞ an = 0

limn→∞ ln(1+an)/an = 1, limn→∞ an = 0

limn→∞ (1-cos(an))/an2 = 1/2, limn→∞ an = 0

limn→∞ n⋅sin(1/n) = ∞⋅0; limn→∞ sin(1/n) = sin(0) = 0; limn→∞ n⋅sin(1/n) = limn→∞ (sin(1/n))/1/n = 1

limn→∞ (sin(n))/n = 0; a bounded sequence, divided by a divergent sequence goes to zero.

limn→∞ (1+1/n)n = 1 = e; limn→∞ 1n = 1 = limn→∞ 1 = 1; indeterminate forms can be resolved.

Cesàro criterion: {an}n∈ℕ | an ≥ 0 ∀ n ∈ ℕ, limn→∞ an+1/an = l ∈ , limn→∞ nan = l

If a sequence has a limit it implies that it is bounded, but if a sequence is bounded it does not imply that it has a limit, like sin(n).

A monotone sequence is always regular, if it is bounded it admits limit, if it is not bounded and increasing it diverges to +∞, if it is not bounded and decreasing it diverges to -∞.

In a sequence defined by recurrence, each element depends on the preceding ones.

In the Fibonacci sequence each number is the sum of the two preceding ones, starting from 0 and 1; a0 = 0, a1 = 1, an = an-1+an-2, n > 1; the beginning of the sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144...; under some older definitions, the value a0 = 0 is omitted, so that the sequence starts with a1 = a2 = 1, and the recurrence an = an−1+an−2 is valid for n > 2; the Fibonacci sequence can also be defined as an+2 = an+1+an, a0 = a1 = 1