If you can’t be kind, at least be vague, Judith Martin
Definition. In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. Sequences are denoted by $\{a_n\}_{n=1}^\infty$
Examples: {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …}, {1, 3, 5, 7, 9, …}, i.e., {2n -1}, {(-1)^{n+1}n} that is 1, −2, 3, −4, 5, −6, 7, −8, 9, −10, . ., etc. The limit of a sequence is the value that the terms of a sequence “tend to”, and is denoted as $\lim_{n \to ∞} a_n = A$. If such a limit exists, the sequence is called convergent. A sequence that does not converge is said to be divergent.
More formally, $\lim_{n \to ∞} {a_n}_{n=1}^\infty = A$ if for all ε > 0, there exists N ∈ ℤ such that ∀n ≥ N: |a_{n} - A| < ε.
Definition. A subsequence of a sequence is a sequence that can be derived or extracted from the original sequence by deleting some of the terms without changing the order of the remaining terms or choosing infinitely many values from the given sequence, where each successive choice is made by taking a larger index (and by doing so, looking farther away in the sequence). Given $\{a_n\}_{n=1}^\infty ⊆ ℝ$,
$\{n_k\}_{k=1}^\infty ⊆ ℕ$,
the sequence $\{a_{n_k}\}_{k=1}^\infty ⊆ ℝ$ is called a subsequence.
Theorem. Every subsequence of a convergent sequence converges and to the same limit.
Proof. Suppose $\{a_n\}_{n=1}^\infty,$
$\lim_{n \to ∞} a_n = A$ and suppose an arbitrary subsequence, $\{a_{n_k}\}_{k=1}^\infty$
Given ε > 0 ⇒ ∃N ∈ ℕ: ∀n ≥ N |a_{n}-L| < ε
By definition of subsequence, $\{n_k\}_{k=1}^\infty ⊆ ℕ$, n_{1} < n_{2} < ··· hence, it is unbounded.
Since {n_{k}} is strictly increasing ⇒ ∃K ∈ ℕ such that ∀k ≥ K ⇒ n_{k} ≥ N ⇒ |a_{nk}-L| < ε ⇒ $\lim_{n_k \to ∞} a_{n_k} = L$ ∎
Bolzano–Weierstrass theorem states that each infinite bounded sequence in ℝ^{n} has a convergent subsequence.
Proof.
Suppose $\{a_n\}_{n=1}^\infty$ is bounded, then |a_{n}| < M for all n ∈ ℕ ↭ a_{n} ∈ [-M, M] = [-M, 0] ∪ [0, M] ⇒ there are infinite-many a_{n} in [-M, 0] ∪ [0, M] ⇒ there are infinite-many a_{n} in [-M, 0] or (maybe there are infinite-many in both) [0, M].
Let’s call it I_{1} the one that has infinite-many a_{n}, and pick an element in $\{a_n\}_{n=1}^\infty ∩ I_1$ and call it a_{n1}.
We iterate this process, let’s split I_{1} into two equal parts, repeat the process and call I_{2} the one that has infinite-many a_{n} and pick an element and call it a_{n2}.
Basically, we are constructing a sequence I_{1} ⊇ I_{2} ⊇ ··· ⊇ I_{k} ⊇ ··· where a_{n1}∈ I_{1}, a_{n2}∈ I_{2}, and so on ⇒[By the nested interval theorem for the real numbers. It states that if you have a sequence of nested, bounded, nonempty, closed intervals, then their intersection is nonempty.] $\bigcap^∞_{i=1} {I_i} ≠ ∅$, length(I_{1}) = M.
Besides, length(I_{2}) = $\frac{M}{2}, ···, length(I_k) = \frac{M}{2^{k-1}}$
$\bigcap^∞{i=1} {I_i} ≠ ∅$, pick an element L ∈ $\bigcap^∞{i=1} {I_i} ≠ ∅$. Claim: $\lim_{k \to ∞} a_{n_k} = L.$
Let ε > 0, we need to choose K, such that ∀k ≥ K ⇒ $|a_{n_k} - L| < ε$
We know that ∀k ≥ K, a_{nk}∈ I_{k} and $length(I_k) = \frac{M}{2^{k-1}}$.
We know that $\lim_{k \to ∞} \frac{1}{2^{k-1}} = 0$. Given ε/M > 0, we choose K > 0, such that ∀k ≥ K, $|\frac{1}{2^{k-1}} - 0|< \frac{ε}{M} ↭ \frac{1}{2^{k-1}} < \frac{ε}{M} ↭ \frac{M}{2^{k-1}}$ <[🚀] ε
In conclusion, let ε > 0, we choose K > 0, such that ∀k ≥ K ⇒ a_{nk}∈ I_{k}, L ∈ I_{k} and $length(I_k) = \frac{M}{2^{k-1}}$ ⇒ $|a_{n_k} - L| < \frac{M}{2^{k-1}}$ <[🚀] ε ∎
However, the subsequence $\{(-1)^{2n}\}_{n=1}^\infty = (1, 1, 1, 1, ···)$ converges to 1.
Besides, $\{(-1)^{2n+1}\}_{n=1}^\infty = (-1, -1, -1, -1, ···)$ converges to -1.
Recall. A continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on.
We could use the continued fraction representation of 2π to find a increasing sequence of integers such that sin(n) approaches sin(2π)=0.
This is how you find the continued fraction for any number at all, say x_{0}.
First, let a_{0} be the largest integer that does not exceed x_{0}, that is, a_{0}=⌊x_{0}⌋. And let b_{0} be the fractional part of x_{0}, that is b_{0} = x_{0} −a_{0}. In our example, x_{0} = π, a_{0} = 3, b_{0} = 0.14159…
Now, we can write the following expression, x _{0} = a_{0} + b_{0} = a_{0} + $\frac{1}{x_1}, x_1 = \frac{1}{b_0} = \frac{1}{0.14159…} = 7.0625···$
Then, repeat, let a_{1}= ⌊x_{1}⌋, b_{1} = x_{1} −a_{2}, and $x_2 = \frac{1}{b_1}$, so we have a_{1} = 7, b_{1} ≈ 0.0625···, and x_{2} = $\frac{1}{0.0625…} = 15.996···$
π = $3 + \frac{1}{7+\frac{1}{15.996···}}$, and in general, x_{0} = $a_0 + \frac{1}{a_1+\frac{1}{x_2}}$
Continued fraction for π: (π = [3;7,15,1,292,1,1,1,2,1,3,1,…])
As it was previously stated, we could use the continued fraction representation of 2π to find a increasing sequence of integers such that sin(n) approaches sin(2π)=0
2·π = 6.283185307179586 = 6 + 0.283185307179586, a_{1} = 6, x_{1} = 1/0.283185307179586 = 3.3531256652965528, a_{2} = 3, x_{2} = 1/3531256652965528
$2\pi = 6 + \frac{1}{3 + \frac{1}{1 + \frac{1}{1 + \frac{1}{7 + \frac{1}{2 + \frac{1}{146 + \frac{1}{3 + \dots}}}}}}}$
First examples of finite simple continuos fraction, a_{0} = 6, $a_0+\frac{1}{a_1} = 6 + \frac{1}{3} = \frac{19}{3}, a_0+\frac{1}{a_1+\frac{1}{a_2}} = 6 + \frac{1}{3+\frac{1}{1}} = 6 + \frac{1}{4} = \frac{25}{4},$ and so on.
The series of 2π is 6, $6, \frac{19}{3}, \frac{25}{4}, \frac{44}{7}, \frac{333}{53}, \frac{710}{113}, \frac{103993}{16551}, \frac{312689}{49766}, \frac{1980127}{315147}, \frac{2292816}{364913}, ···$
Ignoring the denominators (it will not make any difference at infinite), we have the subsequence:
sin(6) = −0.279415, sin(19) = 0.149877, sin(25) = −0.132352, sin(44) =0.0177019, sin(333) = −0.00882117, sin(710) =0.0000602887, sin(103993) = −0.0000191293, sin(312689) = 2.900699×10^{6}, etc. This subsequence is obviously convergent to zero.
$\{2n\}_{n=1}^\infty = (2, 4, 6, 8, ···)$ is unbounded, the sequence may have a convergent subsequence or it may not. In that particular case, the sequence do not have a convergent subsequence.
$\{((-1)^{n}+1)n\}_{n=1}^\infty$ is unbounded, but it has a convergent subsequence, namely
$\{((-1)^{2n+1}+1)(2n+1)\}_{n=1}^\infty= (0, 0, 0, 0, ···)$.