LIMIT OF A SEQUENCE

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

|x| < r ⇔ -r < x < r; |a_{n}-l| < ε ⇔ -ε < a_{n}-l < ε ⇔ l-ε < a_{n} < l+ε; ∀ ε > 0, |a_{n}-l| < ε definitely

lim_{n→+∞} 1/n = 0 ⇔ ∀ ε > 0, ∃ n_{0} | ∀ n > n_{0}, |(1/n)-0| < ε; |1/n| < ε ⇔ -ε < 1/n < ε ⇔ 1/n < ε ⇔ n > 1/ε; n_{0} = [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 lim_{n→∞} a_{n} = 0 then {a_{n}}_{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: {a_{n}}_{n∈ℕ}, l_{1}, l_{2} ∈ ℝ, l_{1} ≠ l_{2} | lim_{n→∞} a_{n} = l_{1} and lim_{n→∞} a_{n} = l_{2}; l_{1} ≠ l_{2} ⇒ |l_{1}-l_{2}| ≠ 0; ε := |l_{1}-l_{2}| > 0; ∃ n_{1} | ∀ n > n_{1}, |a_{n}-l_{1}| < ε/4; ∃ n_{2} | ∀ n > n_{2}, |a_{n}-l_{2}| < ε/4; n_{0} = max{n_{1},n_{2}}; ∀ n > n_{0} = max{n_{1},n_{2}}, ε = |l_{1}-l_{2}| = |l_{1}-a_{n}+a_{n}-l_{2}| [1] ≤ |l_{1}-a_{n}|+|a_{n}-l_{2}| = |a_{n}-l_{1}|+|a_{n}-l_{2}| < (ε/4)+(ε/4) = ε/2 ⇒ ε < ε/2 ⇒ 1 < 1/2 that is a contradiction; [1] Triangle inequality, |x+y| ≤ |x|+|y|

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

lim_{n→∞} n^{2}; {n^{2}}_{n∈ℕ} = {0, 1, 4, 9, ...}

{a_{n}}_{n∈ℕ} diverges to +∞ for n→+∞, lim_{n→+∞} a_{n} = +∞, if ∀ M ∈ ℝ, ∃ n_{0} = n_{0}(M) ∈ ℕ | ∀ n ≥ n_{0} then a_{n} > M

{a_{n}}_{n∈ℕ} diverges to -∞ for n→+∞, lim_{n→+∞} a_{n} = -∞, if ∀ M ∈ ℝ, ∃ n_{0} = n_{0}(M) ∈ ℕ | ∀ n ≥ n_{0} then a_{n} < M

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

lim_{n→+∞} a_{n} = +∞ if ∀ M ∈ ℝ, a_{n} > M definitely

lim_{n→+∞} a_{n} = -∞ if ∀ M ∈ ℝ, a_{n} < M definitely

lim_{n→∞} n^{2} = +∞; ∀ M ∈ ℝ, ∃ n_{0} = n_{0}(M) ∈ ℕ | ∀ n ≥ n_{0}, n^{2} > M; n^{2} > M ⇔ {∀ n ∈ ℕ, M < 0; n > √M, M ≥ 0}; n > √M ⇔ n ≥ [√M]+1 = n_{0}(M)

lim_{n→∞} -n^{3} = -∞

{(-1)^{n}}_{n∈ℕ}

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

{a_{n}} 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}⋅n^{n}; this sequence is irregular or oscillating {0, -1, 4, -27, 256, -3125, ...}

a_{n} = {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; ∄ lim_{n→∞} sin(n)

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

lim_{n→∞} (n^{n}+n!)/(e^{n}+2^{n}) = +∞

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

Harmonic sequence {n^{α}}_{n∈ℕ}, α ∈ ℝ; lim_{n→∞} n^{α} = {+∞, α > 0; 1, α = 0; 0, α < 0}; lim_{n→∞} n^{3/2} = +∞; lim_{n→∞} 1/n^{2} = 0, 1/n^{2} = n^{-2}; lim_{n→∞} n^{0} = lim_{n→∞} 1 = 1

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

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

How can we calculate the limit of a sequence? To apply the definition of limit lim_{n→∞} a_{n} = l, we need to know l ∈ ℝ

The limit of a sum is the sum of the limits: lim_{n→∞} (a_{n}+b_{n}) = l+m; lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = m, l,m ∈ ℝ

The limit of a difference is the difference of the limits: lim_{n→∞} (a_{n}-b_{n}) = l-m; lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = m, l,m ∈ ℝ

The limit of a product is the product of the limits: lim_{n→∞} (a_{n}⋅b_{n}) = l⋅m; lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = m, l,m ∈ ℝ

The limit of a ratio is the ratio of the limits: lim_{n→∞} (a_{n}/b_{n}) = l/m, m ≠ 0; lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = m, l,m ∈ ℝ

The limit of a power is the power of the limits: lim_{n→∞} (a_{n})^b_{n} = l^{m}, l > 0; lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = m, l,m ∈ ℝ

The limit of a module is the module of the limit: lim_{n→∞} |a_{n}| = |l|; lim_{n→∞} a_{n} = l, l ∈ ℝ

lim_{n→∞} (n^{2}-n)/(n^{2}-1) = lim_{n→∞} n^{2}(1-(1/n))/n^{2}(1-(1/n^{2})) = lim_{n→∞} (1-(1/n))/(1-(1/n^{2})) = 1/1 = 1; lim_{n→∞} 1/n = lim_{n→∞} n^{-1} = 0; lim_{n→∞} 1/n^{2} = lim_{n→∞} n^{-2} = 0

lim_{n→∞} a_{n} = l ∈ ℝ, lim_{n→∞} b_{n} = ±∞, lim_{n→∞} a_{n}+b_{n} = ±∞; l±∞ = ±∞

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

l±∞ = ±∞ ∀ l ∈ ℝ

+∞+∞ = +∞; lim_{n→∞} a_{n} = +∞, lim_{n→∞} b_{n} = +∞ ⇒ lim_{n→∞} a_{n}+b_{n} = +∞

-∞-∞ = -∞; lim_{n→∞} a_{n} = -∞, lim_{n→∞} b_{n} = -∞ ⇒ lim_{n→∞} a_{n}+b_{n} = -∞

+∞⋅l = {+∞, l > 0; -∞, l < 0}

l/±∞ = 0 ∀ l ∈ ℝ

q^{+∞} = {+∞, q > 1; 0, q < 1}

+∞⋅+∞ = +∞

-∞⋅-∞ = +∞

+∞⋅-∞ = -∞

-∞⋅+∞ = -∞

lim_{n→∞} (n^{2})+(1/n)+(3^{n}) = +∞+0+∞ = +∞

lim_{n→∞} (n^{2}+2n-1)/(n^{5}+3n+2) = lim_{n→∞} (n^{2}(1+(2/n)-(1/n^{2})))/(n^{5}(1+(3/n^{4})+(2/n^{5}))) = (1+(2/n)-(1/n^{2}))/(n^{3}(1+(3/n^{4})+(2/n^{5}))) = 1/+∞ = 0

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

lim_{n→∞} a_{n} = +∞, lim_{n→∞} b_{n} = -∞, lim_{n→∞} a_{n}+b_{n} = ?

a_{n} = n+1, b_{n} = -n; lim_{n→∞} a_{n} = +∞, lim_{n→∞} b_{n} = -∞ ⇒ lim_{n→∞} a_{n}+b_{n} = lim_{n→∞} 1 = 1

a_{n} = n, b_{n} = -n; lim_{n→∞} a_{n} = +∞, lim_{n→∞} b_{n} = -∞ ⇒ lim_{n→∞} a_{n}+b_{n} = 0

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

a_{n} = n^{3}, b_{n} = n^{2}; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} n^{3}/n^{2} = ∞/∞ = lim_{n→∞} n = +∞; this sequence was initially considered indeterminate, but it turned out to be divergent

a_{n} = 2n^{3}, b_{n} = n^{3}; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} 2n^{3}/n^{3} = ∞/∞ = lim_{n→∞} 2 = 2; this sequence was initially considered indeterminate, but it turned out to be convergent

a_{n} = n, b_{n} = n^{4}; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} n/n^{4} = ∞/∞ = lim_{n→∞} 1/n^{3} = 1/∞ = 0; this sequence was initially considered indeterminate, but it turned out to be infinitesimal

l/0, l ≠ 0, is an indeterminate form

a_{n} = 1, b_{n} = (-1)^{n}/n; lim_{n→∞} a_{n} = lim_{n→∞} 1 = 1; lim_{n→∞} b_{n} = lim_{n→∞} (-1)^{n}/n = 0; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} 1/(-1)^{n}/n = 1/0 = lim_{n→∞} n/(-1)^{n} = lim_{n→∞} (-1)^{n}n = ∄; this sequence was initially considered indeterminate, but it turned out to be irregular

∞^{0} is an indeterminate form; a^{b} = e^{b·ln(a)}; ∞^{0} = e^{0·ln(∞)} = e^{0·∞}

If a_{n} is a bounded sequence, ∃ m,M ∈ ℝ | m ≤ a_{n} ≤ M ∀ n; lim_{n→∞} b_{n} = 0 ⇒ lim_{n→∞} a_{n}b_{n} = 0; a bounded sequence, multiplied by an infinitesimal sequence, is equal to zero

If a_{n} is a bounded sequence, ∃ m,M ∈ ℝ | m ≤ a_{n} ≤ M ∀ n; lim_{n→∞} b_{n} = ±∞ ⇒ lim_{n→∞} a_{n}/b_{n} = 0; a bounded sequence, divided by a divergent sequence, is equal to zero

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

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

Comparison theorem: {a_{n}}_{n∈ℕ}, {b_{n}}_{n∈ℕ} | {a_{n}} ≤ {b_{n}} ∀ n ∈ ℕ, lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = m, l ≤ m

Squeeze theorem or sandwich theorem or theorem of carabinieri: {a_{n}}_{n∈ℕ}, {b_{n}}_{n∈ℕ}, {c_{n}}_{n∈ℕ} | a_{n} ≤ c_{n} ≤ b_{n} ∀ n ∈ ℕ, lim_{n→∞} a_{n} = lim_{n→∞} b_{n} = l, lim_{n→∞} c_{n} = l

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

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

lim_{n→∞} 1/(5^{n}+cos(n)); 5^{n}-1 ≤ 5^{n}+cos(n) ≤ 5^{n}+1; 1/(5^{n}+1) ≤ 1/(5^{n}+cos(n)) ≤ 1/(5^{n}-1); 0 ≤ 1/(5^{n}+cos(n)) ≤ 0; for the squeeze theorem lim_{n→∞} 1/(5^{n}+cos(n)) = 0

lim_{n→∞} ^{n}√a = 1 ∀ a > 0; for a > 1, lim_{n→∞} ^{n}√a = 1 ⇔ lim_{n→∞} ^{n}√a-1 = 0; d_{n} := ^{n}√a-1; we have to prove that lim_{n→∞} d_{n} = 0; d_{n} = ^{n}√a-1 ⇔ ^{n}√a = 1+d_{n} ⇔ a = (1+d_{n})^{n} [1] ≥ 1+nd_{n} ⇔ a ≥ 1+nd_{n} ⇔ a-1 ≥ nd_{n} ⇔ (a-1)/n ≥ d_{n} ≥ 0 [2] ⇒ lim_{n→∞} d_{n} = 0; [1] Bernoulli's inequality, (1+x)^{n} ≥ 1+nx ∀ x > -1, ∀ n ∈ ℕ; [2] for the squeeze theorem; for 0 < a < 1, lim_{n→∞} ^{n}√a = lim_{n→∞} 1/(1/^{n}√a) = lim_{n→∞} 1/^{n}√1/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

{a_{n}}_{n∈ℕ} | lim_{n→∞} a_{n} = 0; if a_{n} ≥ 0 ∀ n ∈ ℕ, then lim_{n→∞} a_{n} = 0^{+}; if a_{n} ≤ 0 ∀ n ∈ ℕ, then lim_{n→∞} a_{n} = 0^{-}

l/0^{+} = {+∞, l > 0; -∞, l < 0}

l/0^{-} = {+∞, l < 0; -∞, l > 0}

lim_{n→∞} a_{n} = l, lim_{n→∞} b_{n} = 0^{+} ⇒ lim_{n→∞} a_{n}/b_{n} = {+∞, l > 0; -∞, l < 0}

lim_{n→∞} 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 {a_{n}}_{n∈ℕ} is monotone, then it is regular, that is convergent or divergent; if {a_{n}}_{n∈ℕ} is increasing, then lim_{n→∞} a_{n} = sup{a_{n} | n ∈ ℕ}; if {a_{n}}_{n∈ℕ} is decreasing, then lim_{n→∞} a_{n} = inf{a_{n} | 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{a_{n} | n ∈ ℕ} ∈ ℝ; thesis l = lim_{n→∞} a_{n} ⇔ ∀ ε > 0, ∃ n_{0} | ∀ n > n_{0}, |a_{n}-l| < ε ⇔ -ε < a_{n}-l < ε; by definition of supremum, l is a majorant or upper bound of {a_{n} | n ∈ ℕ}, therefore a_{n} ≤ l ∀ n ∈ ℕ ⇔ a_{n}-l ≤ 0 < ε ∀ n ∈ ℕ; by definition of supremum, l-ε is not a majorant or upper bound of {a_{n} | n ∈ ℕ} ⇒ ∃ n_{0} ∈ ℕ | l-ε < a_{n0}, because {a_{n}}_{n∈ℕ} is increasing, a_{n} ≥ a_{n0} ∀ n > n_{0}, therefore -ε ≤ a_{n}-l ∀ n ≥ n_{0}

{(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

lim_{n→∞} (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

{a_{n}}_{n∈ℕ}, {b_{n}}_{n∈ℕ} are asymptotic sequences if lim_{n→∞} a_{n}/b_{n} = 1; a_{n} ~ b_{n} n→∞

a_{n} ~ b_{n} ⇒ lim_{n→∞} a_{n} = lim_{n→∞} b_{n}

a_{n} = n^{2}-3n+2, b_{n} = n^{2}+1; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} (n^{2}-3n+2)/(n^{2}+1) = lim_{n→∞} n^{2}(1-3/n+2/n^{2})/n^{2}(1+1/n^{2}) = lim_{n→∞} (1-3/n+2/n^{2})/(1+1/n^{2}) = 1/1 = 1 ⇒ n^{2}-3n+2 ~ n^{2}+1

a_{n} = n^{5/2}+n^{2}+n^{1/2}+2, b_{n} = n^{5/2}; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} (n^{5/2}+n^{2}+n^{1/2}+2)/(n^{5/2}) = lim_{n→∞} 1+1/n^{1/2}+1/n^{2}+2/n^{5/2} = 1

lim_{n→∞} a_{n} = lim_{n→∞} b_{n} !⇒ a_{n} ~ b_{n}

a_{n} = n^{2}, b_{n} = n; lim_{n→∞} a_{n} = lim_{n→∞} b_{n} = +∞; lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} n^{2}/n = +∞

The asymptotic relation is an equivalence relation: it is reflexive, symmetrical, and transitive; reflexive: a_{n} ~ a_{n} ⇔ lim_{n→∞} a_{n}/a_{n} = 1; symmetrical: a_{n} ~ b_{n} ⇔ b_{n} ~ a_{n}, lim_{n→∞} b_{n}/a_{n} = lim_{n→∞} 1/a_{n}/b_{n} = lim_{n→∞} 1/1 = 1; transitive: a_{n} ~ b_{n}, b_{n} ~ c_{n} ⇒ a_{n} ~ c_{n}, lim_{n→∞} a_{n}/c_{n} = lim_{n→∞} (a_{n}/b_{n})(b_{n}/c_{n}) = lim_{n→∞} 1⋅1 = 1

Substitution principle: if a_{n} ~ a_{n}^{'}, b_{n} ~ b_{n}^{'}, c_{n} ~ c_{n}^{'}, then lim_{n→∞} a_{n}⋅b_{n}/c_{n} = lim_{n→∞} a_{n}^{'}⋅b_{n}^{'}/c_{n}^{'}

lim_{n→∞} (n^{3}+2n+1)/(n^{4}+4n^{2}+3); n^{3}+2n+1 ~ n^{3}, n^{4}+4n^{2}+3 ~ n^{4}; lim_{n→∞} (n^{3}+2n+1)/(n^{4}+4n^{2}+3) = lim_{n→∞} n^{3}/n^{4} = lim_{n→∞} 1/n = 0

The substitution principle applies only to products and ratios: a_{n} ~ a_{n}^{'}, b_{n} ~ b_{n}^{'} ⇒ lim_{n→∞} a_{n}/b_{n} = lim_{n→∞} a_{n}^{'}/b_{n}^{'}, lim_{n→∞} a_{n}b_{n} = lim_{n→∞} a_{n}^{'}⋅b_{n}^{'}

The substitution principle does not generally apply to sums, differences and powers: a_{n} ~ a_{n}^{'}, b_{n} ~ b_{n}^{'} !⇒ lim_{n→∞} a_{n}±b_{n} = lim_{n→∞} a_{n}^{'}±b_{n}^{'}, lim_{n→∞} a_{n}^b_{n} = lim_{n→∞} a_{n}^{'}^b_{n}^{'}

a_{n} = n+1 ~ a_{n}^{'} = n, b_{n} = -n ~ b_{n}^{'} = -n; lim_{n→∞} a_{n}+b_{n} = lim_{n→∞} n+1-n = 1; lim_{n→∞} a_{n}^{'}+b_{n}^{'} = lim_{n→∞} n-n = 0

lim_{n→∞} a_{n} = l ⇔ lim_{n→∞} a_{n}/l = 1; a_{n} ~ l

a_{n} = 1+1/n ~ a_{n}^{'} = 1, b_{n} = n ~ b_{n}^{'} = n; lim_{n→∞} a_{n}^b_{n} = lim_{n→∞} (1+1/n)^{n} = e; lim_{n→∞} a_{n}^{'}^b_{n}^{'} = lim_{n→∞} 1^{n} = lim_{n→∞} 1 = 1

lim_{n→∞} ln(n)/n^{α} = 0, α > 0; it would be ∞/∞, but the harmonic sequence at the denominator grows faster than the logarithm sequence at the numerator

lim_{n→∞} n^{α}/q^{n} = 0, α > 0, q > 1; it would be ∞/∞, but the geometric sequence at the denominator grows faster than the harmonic sequence at the numerator

lim_{n→∞} q^{n}/n! = 0, q > 1; it would be ∞/∞, but the factorial sequence at the denominator grows faster than the geometric sequence at the numerator

lim_{n→∞} n!/n^{n} = 0; it would be ∞/∞, but the n^{n} sequence at the denominator grows faster than the factorial sequence at the numerator

Infinity hierarchy from slowest to fastest: ln(n), n^{α}, q^{n}, n!, n^{n}

lim_{n→∞} q^{n}/n^{n} = lim_{n→∞} (q^{n}/n!)(n!/n^{n}) = 0

lim_{n→∞} n^{n}/q^{n} = lim_{n→∞} 1/q^{n}/n^{n} = 1/0^{+} = +∞

lim_{n→∞} ^{n}√n = 1; a^{b} = e^{b⋅ln(a)}; lim_{n→∞} ^{n}√n = lim_{n→∞} n^{1/n} = lim_{n→∞} e^{(ln(n))/n} = lim_{n→∞} e^{0} = 1

lim_{n→∞} (1+x/a_{n})^{an} = e^{x}, ∀ x ∈ ℝ, ∀ {a_{n}}_{n∈ℕ} | lim_{n→∞} a_{n} = +∞; lim_{n→∞} (1+1/n)^{n} = e, x = 1, a_{n} = n; lim_{n→∞} (1+2/(n^{2}+1))^{n2+1} = e^{2}, x = 2, a_{n} = n^{2}+1

lim_{n→∞} sin(a_{n})/a_{n} = 1, lim_{n→∞} a_{n} = 0

lim_{n→∞} (e^{an}-1)/a_{n} = 1, lim_{n→∞} a_{n} = 0

lim_{n→∞} ln(1+a_{n})/a_{n} = 1, lim_{n→∞} a_{n} = 0

lim_{n→∞} (1-cos(a_{n}))/a_{n}^{2} = 1/2, lim_{n→∞} a_{n} = 0

lim_{n→∞} n⋅sin(1/n) = ∞⋅0; lim_{n→∞} sin(1/n) = sin(0) = 0; lim_{n→∞} n⋅sin(1/n) = lim_{n→∞} (sin(1/n))/1/n = 1

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

lim_{n→∞} (1+1/n)^{n} = 1^{∞} = e; lim_{n→∞} 1^{n} = 1^{∞} = lim_{n→∞} 1 = 1; indeterminate forms can be resolved

Cesàro criterion: {a_{n}}_{n∈ℕ} | a_{n} ≥ 0 ∀ n ∈ ℕ, lim_{n→∞} a_{n+1}/a_{n} = l ∈ ℝ, lim_{n→∞} ^{n}√a_{n} = 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; a_{0} = 0, a_{1} = 1, a_{n} = a_{n-1}+a_{n-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 a_{0} = 0 is omitted, so that the sequence starts with a_{1} = a_{2} = 1, and the recurrence a_{n} = a_{n−1}+a_{n−2} is valid for n > 2; the Fibonacci sequence can also be defined as a_{n+2} = a_{n+1}+a_{n}, a_{0} = a_{1} = 1