Notation: We write \((a_n)_n\) as shorthand for \((a_n)_{n = 1\vert 0}^\infty\) to avoid specifying whether the seuqnece begins at 1 or 0.
Definition: We say that a sequence \((a_n)_n\) converges to \(L \in \mathbb R\) denoted \(\lim_{n \to \infty} a_n = L\) when
\[\begin{align*} \forall \epsilon \gt 0, \exists N \in \mathbb N, \forall n \geq N, \lvert a_n - L \rvert \lt \epsilon \end{align*}\]When there is no such \(L\), the sequence diverges.
Definition: We say \((a_n)_n\) diverges to \(\infty\), \(\lim_{n \to \infty} a_n = \infty\) when
\[\forall M \in \mathbb R, \exists N \in \mathbb N, \forall n \geq N, a_n \gt M\]and diverges to \(-\infty\) when \(\lim_{n \to \infty}a_n = -\infty\) when
\[\begin{align*} \forall m \in \mathbb R, \exists N \in \mathbb N, \forall n \geq N, m \lt a_n \end{align*}\]Definition: A sequence is bounded when \(\exists M \gt 0, \forall n, \vert a_n \vert \leq M\).
Theorem:
Proof:
Theorem: If \(\lim_{n \to \infty}a_n\) exists, then it is unique.
Proof: Supppose \((a_n)_n\) such that \(L_1, L_2\) satisfy the definition of \(\lim_{n \to \infty}a_n\) and WLOG, \(L_1 \lt L_2\). Choose \(\epsilon = \frac{L_2 - L_1}{2}\). Choose \(N_1 \in \mathbb N\) such that \(\forall n \geq N_1\), \(\lvert a_n - L_1 \rvert \lt \epsilon\). Similarily, choose \(N_2 \in \mathbb N, \forall n \geq N_2, \lvert a_n - L_2 \rvert \lt \epsilon\). Choose \(N_0 = \max \lbrace N_1, N_2 \rbrace\). Thus, \(N_0 \geq N_1\) and \(N_0 \geq N_2\). Then,
\(\begin{align*} \lvert L_1 - L_2 \rvert &= \lvert (L_1 - a_{N_0}) + (a_{N_0} - L_2) \rvert \\ &\leq \lvert L_1 - a_{N_0} \rvert + \lvert a_{N_0} - L_2 \rvert \\ &\lt \epsilon + \epsilon \\ &= \frac{2 \lvert L_1 - L_2 \rvert}{2} \\ &= \lvert L_1 - L_2 \rvert \end{align*}\) So \(\vert L_1 - L_2 \vert \lt \vert L_1 - L_2 \vert\); a contradiction. \(\blacksquare\)
Theorem (Algebraic Limit): Given convergent sequences \((a_n)_n, (b_n)_n\) and \(c \in \mathbb R\), the following hold:
Proof: Let \(\lim a_n = A, \lim b_n = B\).
(1) We want to show that \(\forall \epsilon \gt 0, \exists N \in \mathbb N, \forall n \geq N, \lvert (a_n + b_n) - (A + B) \rvert \lt \epsilon\). Let \(\epsilon \gt 0\). Choose \(N_1 \in \mathbb N\) such that \(\forall n \geq N_1, \lvert a_n - A \rvert \lt \epsilon/2\). Likewise, choose \(N_2 \in \mathbb N\) such that \(\forall n \geq N_2, \lvert b_n - B \rvert \lt \epsilon/2\). Define \(N_0 = \max \lbrace N_1, N_2 \rbrace\). Then, \(\forall n \geq N_0\), \(n \geq N_1\) and \(n \geq N_2\). So,
\[\begin{align*} \lvert (a_n + b_n) - (A + B) \rvert \leq \lvert a_n - A \rvert + \lvert b_n - B \rvert \lt \epsilon/2 + \epsilon/2 = \epsilon \end{align*}\]$\blacksquare$
(2) Let \(\epsilon \gt 0\). Choose \(N_1, N_2 \in \mathbb N\) such that \(\forall n \geq N_1, \vert a_n - A \vert \lt \epsilon/2\) and \(\forall n \geq N_2, \vert b_n - B \vert \lt \epsilon/2\). Define \(N_0 = \max \lbrace N_1, N_2 \rbrace\). Then, \(\forall n \geq N_0, n \geq N_1\) and \(n \geq N_2\). So,
\[\begin{align*} \vert (a_n - b_n) - (A - B) \vert \leq \vert a_n - A \vert + \vert B - b_n \vert \lt \epsilon/2 + \epsilon/2 = \epsilon \end{align*}\]$\blacksquare$
(3) Let \(\epsilon \gt 0\). Choose \(N \in \mathbb N\) such that \(\forall n \geq N\), \(\vert a_n - A \vert \lt \epsilon/c\). Then,
\[\begin{align*} c \vert a_n - A \vert \lt c \cdot \epsilon/c \implies \vert ca_n - cA \vert \lt \epsilon. \end{align*}\]$\blacksquare$
(4) We want to show \(\forall \epsilon \gt 0, \exists N_0 \in \mathbb N, \forall n \geq _0, \lvert a_n b_n - AB \rvert \lt \epsilon\). Observe
\[\begin{align*} \lvert (a_n b_n - a_n B) + (a_n B - AB) \rvert \leq \lvert a_n \rvert \lvert b_n - B \rvert + \lvert B \rvert \lvert a_n - A \rvert \end{align*}\]Since \((a_n)_n\) is convergent, it is bounded, so we can choose an upper bound \(M \gt 0\) such that \(\lvert a_n \rvert \leq M\) for all \(n \geq 1\).
Given an arbitrary \(\epsilon \gt 0\), choose \(N_1, N_2 \in \mathbb N\) such that
\[\begin{align*} \lvert a_n - A \rvert &\lt \frac{\epsilon}{M + \lvert B \rvert}, \forall n \geq N_1 \\ \lvert b_n - B \rvert &\lt \frac{\epsilon}{M + \lvert B \rvert}, \forall n \geq N_2 \end{align*}\]Now, set \(N_0 = \max \lbrace N_1, N_2 \rbrace\). Then, \(\forall n \geq N_0\),
\[\begin{align*} \lvert a_nb_n - AB \rvert &\leq \lvert a_n \rvert \cdot \lvert b_n - B \rvert + \lvert B \rvert \cdot \lvert a_n - A \rvert \\ &\leq M \frac{\epsilon}{M + \lvert B \rvert} + \lvert B \rvert \frac{\epsilon}{M + \lvert B \rvert} \\ &= \epsilon \end{align*}\]$\blacksquare$
(5) Observe
\[\begin{align*} \lvert \frac{a_n}{b_n} - \frac{A}{B} \rvert &= \lvert \frac{a_nB - b_nA}{b_nB} \rvert \\ &= \lvert \frac{(a_nB - AB) + (AB - b_nA)}{b_nB} \rvert \\ &\leq \frac{\lvert a_n - A \rvert \lvert B \rvert}{\lvert b_n \rvert \lvert B \rvert} + \frac{\lvert A \rvert \lvert b_n - B \rvert}{\lvert b_n \rvert \lvert B \rvert} \end{align*}\]Since \(B \neq 0\), there exists \(N_1 \in \mathbb N\) such that \(\vert b_n - B \vert \lt \vert B \vert/2\) for all \(n \geq N_1\), so \(\vert b_n \gt \vert B \vert / 2\) for all \(n \geq N_1\). Let \(\epsilon \gt 0\). TODO.
Theorem: Let \((a_n)_n, (b_n)_n\) be sequences, \(A \in \mathbb R\), \(r \gt 0\), and \(\lim_{n \to \infty} b_n = 0\). If \(\vert a_n - A \vert \leq r \cdot \vert b_n \vert\) for all but finitely many \(n \in \mathbb N\), then \(\lim_{n \to \infty} a_n = A\).
Proof: Given \(\epsilon \gt 0\), let \(N_0 \in \mathbb N\) be such that \(\forall n \geq N_0\), \(\vert b_n \vert = \vert b_n - 0 \vert \lt \epsilon/r\). Then, \(\forall n \geq N_0, \vert a_n - A \vert \lt r \cdot \epsilon/r = \epsilon\). \(\blacksquare\)
Theorem: Let \((a_n)_n, (b_n)_n\) be sequences with \(\lim_{n \to \infty} a_n = A, \lim_{n \to \infty} b_n = B\).
Proof:
Example: Note that it is not true that \(A \leq B \implies a-n \leq b_n\). For example, consider \(a_n = \frac{(-1)^n}{n}, b_n = 0\).
Theorem (Squeeze Theorem): Given sequences \((a_n), (b_n), (c_n)\) such that \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n\), and, for all but finitely many \(n\), \(a_n \leq b_n \leq c_n\), it follows that \(\lim_{n \to \infty} b_n = \lim_{n \to \infty} a_n\).
Proof: Suppose first that \(\lim_\infty a_n = \lim_\infty c_n = L \in \mathbb R\) and let \(N_1 \in \mathbb N\) such that \(a_n \leq b_n \leq c_n, \forall n \geq N_1\). By the above theorem, since \(a_n \leq b_n \leq c_n, L \leq \lim_{n \to \infty} b_n \leq L\), so \(\lim_{n \to \infty} b_n = L\). \(\blacksquare\)
Lemma (Bernoulli’s Inequality): If \(a \geq -1\), then \((1 + a)^n \geq 1 + na\), \(\forall n \in \mathbb N\).
Proof: By induction on \(n\). We have \((1 + a)^0 = 1 \geq 1 + 0\). For \(k \geq 0\), \((1 + a)^{k + 1} = (1 + a)^k (1 + a) \geq (1 + ka)(1 + a) = 1 + a + ka + ka^2 \geq 1 + (k+1)a\). \(\blacksquare\)
Theorem: Let \(q \in \mathbb R\). Then,
\[\lim_{n \to \infty} q^n = \begin{cases} \infty & q \gt 1 \\ 1 & q = 1 \\ 0 & \vert q \vert \lt 1 \end{cases}\]Proof: If \(q \gt 1\), then \(q^n = (1 + (q - 1))^n \geq 1 + n(q - 1)\). Also, \(\lim_{n \to \infty} cn = \infty\) for every \(c \gt 0\), hence the claim follows from Squeeze Theorem. If instead \(q = 0\) or \(q = 1\), then \((q^n)\) is a constant sequence.
So suppose that \(q \in (0, 1)\). Then, \(1/q \gt 1\) by the axioms of ordred fields, and hence \(1/q^n = (1/q)^n \geq 1 + n(1/q - 1) = 1 + n(\frac{1 - q}{q})\).
Let \(\epsilon \gt 0\) and choose \(N \in \mathbb N\) such that \(1 + N (\frac{1 - q}{q}) \gt 1/\epsilon\). Then \(\forall n \geq N\), \(\frac{1}{q^n} \geq 1 + n(\frac{1 - q}{q}) \geq 1 + N(\frac{1 - q}{q}) \gt \frac{1}{\epsilon}\), hence \(0 \lt q^n \lt \epsilon\), and so \(\vert q^n - 0 \vert \lt \epsilon\).
Finally, suppose \(q \in (-1, 0)\). Then, by above, \(\lim_{n \to \infty} \vert q^n \vert = \lim_{n \to \infty} \vert q \vert^n = 0\), and hence \(\lim_{n \to \infty} q^n = 0\), by the lemma below. \(\blacksquare\).
Lemma: For any sequence \((a_n)\), \(\lim_{n \to \infty} a_n = 0 \iff \lim_{n \to \infty} \vert a_n \vert = 0\).
Proof: Suppose \(\lim_{n \to \infty} a_n = 0\). Let \(\epsilon \gt 0\) and choose \(N \in \mathbb N\) such that \(\forall n \geq N\), \(\vert a_n - 0 \vert \lt \epsilon\). Then, \(\forall n \gt N\),
\[\begin{align*} \vert \vert a_n \vert - 0 \vert = \vert \vert a_n \vert \vert = \vert a_n \vert = \vert a_n - 0 \vert \lt \epsilon. \end{align*}\]Conversely, suppose \(\lim_{n \to \infty} \vert a_n \vert = 0\). Let \(\epsilon \gt 0\) and choose \(N \in \mathbb N\) such that \(\forall n \geq N\), \(\vert \vert a_n \vert - 0 \vert \lt \epsilon\). Then, \(\forall n \gt N\),
\[\begin{align*} \vert a_n - 0 \vert = \vert a_n \vert = \vert \vert a_n \vert \vert = \vert \vert a_n \vert - 0 \vert \lt \epsilon. \end{align*}\] \[\blacksquare\]Theorem:
Proof: TODO.
Theorem: Given a sequence \((a_n)_n\) with non-zero terms, suppose \(\vert \frac{a_n + 1}{a_n} \vert \leq q\) for some \(q \lt 1\) for all but finitely many \(n\). Then, \(\lim_{n \to \infty} a_n = 0\).
Proof: TOOD.
Corollary: Given \((a_n)_n\), if \(\lim_{n \to \infty} \frac{a_{n + 1}}{a_n} = c\) and \(\vert c \vert \lt 1\), then \(\lim_{n \to \infty} a_n = 0\).
Definition: An infinite subsequence of a sequence \((a_n)_n\) is a composition of functions \((a_n) \circ \phi\), where \(\phi: \mathbb N_+ \hookrightarrow \mathbb N_+\) is strictly increasing; written as \((a_{n_k})_k\).
Theorem: Let \((a_n)_n\) be a sequence.
Proof: TODO.
Corollary: Suppose \((a_n)_n\) has two convergent subsequences \((a_{n_k})_k, (a_{n_l})_l\) such that \(\lim_{k \to \infty} a_{n_k} \neq \lim_{l \to \infty} a_{n_l}\). Then, \((a_n)_n\) is divergent.
Definition: A sequence \((a_n)_n\) is called
Theorem (Monotone Convergence): Given \((a_n)\), if \((a_n)\) is increasing and bounded above, or decreasing and bounded below, then \(\exists L \in \mathbb R, \lim_{n \to \infty} a_n = L\).
Proof: Suppose \((a_n)_n\) is increasing and let \(M \gt 0\) be such that \(a_n \leq M, \forall n\). Then, \(A = \lbrace a_n : n \in \mathbb N \rbrace \subseteq \mathbb R\) is non-empty and bounded above. Set \(\alpha = \sup A\). We claim \(\lim_{n \to \infty} a_n = \alpha\). Let \(\epsilon \gt 0\) be arbitrary. Then, \(\alpha - \epsilon\) is not an upper bound of \(A\), so we can choose \(a_{N_0} \gt \alpha - \epsilon\). Then, \(\forall n \geq N_0\), \(\alpha - \epsilon \lt a_{N_0} \leq a_n \leq \alpha \lt \alpha + \epsilon\). That is, \(\lvert a_n - \alpha \rvert \lt \epsilon\). \(\blacksquare\)
Theorem (Bolzano-Weierstrass): If \((s_n)_n\) is bounded, then there is a subsequence \((s_{n_k})_k\) of \((s_n)_n\) where \(\exists L \in \mathbb R\) such that \(\lim_{k \to \infty} s_{n_k} = L\)
Proof: Suppose \(a_1, b_1 \in \mathbb R\) are such that \(a_1 \leq s_n \leq b_1\). Let \(I_1 = [a_1, b_1]\). If \(\lvert \lbrace s_n : n \geq 1 \rbrace \cap (-\infty, \frac{a_1 + b_1}{2}) \rvert = \aleph_0\), (if the first half of the interval is infinite), then we set \(a_2 = a_1, b_2 = \frac{a_1 + b_1}{2}\). Otherwise, we set \(a_2 = \frac{a_1 + b_1}{2}\) and \(b_2 = b_1\), which must be infinite since the sequence is infinite, so it cannot be the union of two finite sets. Set \(I_2 = [a_2, b_2]\). Having constructed \(I_1 \supseteq I_2 \supseteq \dots \supseteq I_k = [a_k, b_k]\), if \(\lvert \lbrace s_n : n \geq 1 \rbrace [a_k, \frac{a_k + b_k}{2}] \rvert = \aleph_0\), then set \(a_{k + 1} = a_k\), \(b_{k + 1} = \frac{a_k + b_k}{2}\), otherwise set \(a_{k + 1} = \frac{a_k + b_k}{2}\), \(b_{k + 1} = b_k\). Set \(I_{k + 1} = [a_{k + 1}, b_{k + 1}]\). By the Nested Interval Property, \(\exists L \in \mathbb R\) such that \(L \in \bigcap_{k = 1}^\infty I_k\).
The construction of the subsequence \((s_{n_k})\) is as follows. Let \(s_{n_1} = s_{n}\). Having chosen \(\lbrace s_{n_1}, \dots, s_{n_k} \rbrace\), we choose \(s_{n_{k + 1}}\) from among \(\lbrace s_n : n \geq 1 \rbrace \cap I_{k + 1} \setminus \lbrace s_{n_1}, \dots, s_{n_k} \rbrace\). Then \((s_{n_k}) \subseteq (s_n)_n\) is such that \(s_{n_k} \in I_k, \forall k \geq 1\).
We claim \(\lim_{k \to \infty} s_{n_k} = L\). Let \(\epsilon \gt 0\) be arbitrary. We have \(\forall k, \operatorname{diam}(I_k) = \frac{b_1 - a_1}{2^{k - 1}}\). Since \(\left( \frac{1}{2} \right)^{k - 1} \to_{k \to \infty} 0\), we can choose \(K_0 \in \mathbb N\) such that \(\operatorname{diam}(I_{K_0}) \lt \epsilon\). Then, \(\forall k \geq K_0\), \(\lvert s_{n_k} - L \rvert \leq \operatorname{diam}(I_k) \leq \operatorname{diam}(I_{K_0}) \lt \epsilon\). \(\blacksquare\)
Definition: A sequence \((a_n)_n\) is Cauchy when \(\forall \epsilon \gt 0, \exists N \in \mathbb N, \forall m, n \geq N, \lvert a_m - a_n \rvert \lt \epsilon\).
Theorem: Let \((a_n)_n\) be a sequence. Then, \((a_n)\) is a Cauchy sequence iff \((a_n)_n\) is convergent.
Proof: (\(\leftarrow\)) Suppose \(\lim_{n \to \infty}a_n = L\). Let \(\epsilon \gt 0\) be arbitrary. We can choose \(N_0\) such that \(\forall n \geq N_0\), \(\lvert a_n - L \rvert \lt \epsilon/2\). Then, for all \(m, n \geq N_0\),
\[\begin{align*} \lvert a_m - a_n \rvert &= \lvert a_m - L + L - a_n \rvert \\ &= \lvert (a_m - L) - (a_n - L) \rvert \\ &\leq \lvert a_m - L \rvert + \lvert a_n - L \rvert \\ &\leq \epsilon/2 + \epsilon/2 \\ &= \epsilon \end{align*}\](\(\rightarrow\)) First we show \(\lbrace a_n : n \geq 1 \rbrace = A\) is bounded. Choose \(N_0 \in \mathbb N\) such that \(\forall n \geq N, \lvert a_{N_0} - a_n \rvert \lt 1\). Hence, \(\forall n \geq 1, \lvert a_n \rvert \leq \max \lbrace \lvert a_1 \rvert, \dots, \lvert a_{N_0 - 1} \rvert, \lvert a_{N_0} \rvert + 1 \rbrace\). So \(A\) is bounded. Now, choose a convergent subsequence \((a_{n_k})_k\) of \((a_n)_n\). Let \(L = \lim_{k \to \infty} a_{n_k}\). We claim that \(\lim_{n \to \infty} a_n = L\). Let \(\epsilon \gt 0\). By the Cauchy property, choose \(N_0 \in \mathbb N\) such that \(\forall m, n \geq N_0\), \(\lvert a_m - a_n \rvert \lt \epsilon/3\). Choose \(K_0 \in \mathbb N\) such that \(\forall k \geq K_0, \lvert a_{n_k} - L \rvert = \epsilon/3\). Set \(N_1 = \max \lbrace N_0, n_{K_0}\rbrace\). Then \(\forall n \geq \mathbb N\), \(\lvert a_n - L \rvert = \lvert a_n - a_{N_{K_0}} + a_{N_{K_0}} - L \rvert\) \(\lt \epsilon/3 + \epsilon/3 \lt \epsilon\). \(\blacksquare\).
Definition: A series \(\sum_{n = 0}^\infty a_n\) is a pair of sequences \(((a_n)_n, (s_n)_n)\) where \((a_n)_{n = 0}^\infty\) is the sequence of terms of the series, and the sequence of partial sums \((s_n)_n\) satisfies \(s_k = a_0 + \dots + a_k = \sum_k a_n\), for all \(k \in \mathbb N\). We say that \(\sum_0^\infty a_n\) is convergent when \(\lim_{n \to \infty} s_n\) exists. If \(\lim_{n \to \infty} s_n = L \in \mathbb R\), we write \(\sum_{n = 0}^\infty a_n = L\) and say that \(L\) is the sum of the series. Otherwise, we say \(\sum a_n\) diverges. If \(\lim_{n \to \infty} s_n = \infty\), we write \(\sum a_n = \infty\) and say the series diverges to \(\infty\).
Theorem (Divergence Test): If the series \(\sum_{n = 0}^\infty a_n\) converges, then \(\lim_{n \to \infty} a_n = 0\).
Proof: By definition, \(a_n = s_n - s_{n - 1} \to_{n \to \infty} L - L = 0\). \(\blacksquare\)
Example: Consider \(\sum_{n = 0}^\infty (-1)^n \frac{n}{2n + 1}\). We have \(\vert a_n \vert = \frac{n}{2n + 1} = \frac{1}{2 + \frac{1}{n}} \to_{n \to \infty} \frac{1}{2} \neq 0\). So \(\lim_{n \to \infty}a_n \neq 0\), thus the series diverges, by the Divergence Test.
Note that the converse of the Divergence test does not hold!
Example 2: The harmonic series \(\sum_{n = 1}^\infty \frac{1}{n} = \infty\), but \(\lim_{n \to \infty}\frac{1}{n} = 0\).
Proof: We claim that a sequence diverges if and only if it has a subsequence that diverges. Suppose a sequence diverges. Then, the subsequence corresponding to the sequence itself diverges. Conversely, suppose there exists a subsequence that diverges. Then, by the above theorem that a sequence converges if and only if every one of its subsequences converges, the sequence must diverge.
So it suffices to find a subsequence \((s_{n_k})_{k=0}^\infty\) of \((s_n)_{n=1}^\infty\) such that \(\lim_{k \to \infty} s_{n_k} = \infty\). Consider \((s_{2^k})_{k = 0}^\infty\). Observe \(\begin{align*} s_1 &= 1 \\ s_2 &= 1 + 1/2 \\ s_4 &= 1 + 1/2 + (1/3 + 1/4) > 1 + 2 \cdot \frac{1}{2} \\ s_8 &= 1 + 1/2 + (1/3 + 1/4) + (1/5 + \dots + 1/8) > 1 + 3 \cdot 1/2 \\ &\dots \\ s_{2^k} &\gt 1 + k \cdot \frac{1}{2} \end{align*}\)
Hence, since \(\lim_{k \to \infty}(1 + k/2) = \infty\), \(\lim_{k \to \infty} s_{n_k} = \infty\). \(\blacksquare\)
Theorem (Cauchy Criterion for Convergence): For a series \(\sum_{n = 0}^\infty a_n\), the following are equivalent:
Proof: By a theorem about sequences, \((s_n)_{n = 0}^\infty\) is convergent iff \((s_n)\) is Cauchy. That is, \(\forall \epsilon \gt 0, \exists N \in \mathbb N, \forall n \gt m \geq N, \vert s_n - s_m \vert \lt \epsilon\). \(\blacksquare\)
Definition: Given a series \(\sum_{n = 0}^\infty\), we say the series converges absolutely when \(\sum_{n = 0}^\infty \vert a_n \vert\) converges. We say \(\sum a_n\) converges conditionally when \(\sum a_n\) converges and \(\sum \vert a_n \vert\) diverges.
Theorem: If \(\sum_{n = 0}^\infty \vert a_n \vert\) converges, then so does \(\sum_{n = 0}^\infty a_n\).
Proof: Given \(\sum_{n = 0}^\infty a_n\) such that \(\sum_{n = 0}^\infty \vert a_n \vert\) is convergent, let \(\epsilon \gt 0\) be arbitrary. Since \(\sum \vert a_n \vert\) satisfies the Cauchy condition, we can choose \(N_0 \in \mathbb N\) such that \(\forall n \gt m \geq N_0, \left\vert \sum_{k = m + 1}^n \vert a_k \vert \right\vert \lt \epsilon\). Then
\[\begin{align*} \left\vert \sum_{k = m + 1}^n a_k \right\vert &\leq \left\vert a_{m + 1} \right\vert + \dots + \vert a_n \vert \\ &= \sum_{k = m + 1}^n \vert a_k \vert \\ &= \left\vert \sum_{k = m + 1}^n \vert a_k \vert \right\vert \\ &\lt \epsilon \end{align*}\]\(\blacksquare\).
Example 3: Consider \(\sum_{n = 1}^\infty \frac{(-1)^n}{n}\). We have \(\vert a_n \vert = \frac{1}{n}\), so \(\sum a_n\) is not absolutely convergent. But \(\sum_1^\infty \frac{(-1)^n}{n}\) converges, by the next theorem, and so \(\sum a_n\) is conditionally convergent.
Theorem (Geometric Series): Let \(q \in \mathbb R\). If \(\vert q \vert \lt 1\), then \(\sum_{n = 0}^\infty q^n\) is absolutely convergent and \(\sum_{n = 0}^\infty q^n = \frac{1}{1 - q}\). If \(\vert q \vert \geq 1\), then \(\sum_n q^n\) is divergent.
Proof: TODO.
Theorem (Comparison Test): Given two series \(\sum_n a_n, \sum _n b_n\) satisfying \(0 \leq a \lt b_n\) for all but finitely many \(n\),
Proof: TODO.
Theorem (Algebraic Convergence Theorem): Given series \(\sum_n a_n, \sum_n b_n\) and a constant \(c \in \mathbb R\), suppose \(\sum_n a_n\) and \(\sum_n b_n\) converge. Then
Proof: Follows immediately from the Algebraic Limit Theorem applied to sequences of the partial sums. \(\blacksquare\).
Theorem (Alternating Series): Suppose \(\sum_{n = 0}^\infty (-1)^n b_n\) is a series such that the sequence \((b_n)_n\) satisfies:
Then \(\sum_0^\infty (-1)^n b_n\) is convergent.
Proof: Choose \(N_0 \in \mathbb N\) such that \(\forall n \geq N_0, b_n \geq 0\) and \(b_{n + 1} \leq b_n\). Consider the subequences \((s_{2m})_{m = 0}^\infty\) and \((s_{2m + 1})_{m = 0}^\infty\). For every \(n \gt m \geq N_0/2\),
\[\begin{align*} s_{2n} &= s_{2n - 1} - (b_{2n - 1} - b_{2n})\\ &= \dots \\ &= s_{2m} - (b_{2m + 1} - b_{2m + 2}) - \dots - (b_{2n - 1} - b_{2n}) \\ &\leq s_{2m} \end{align*}\]so \((s_{2m})_{m = 0}^\infty\) is decreasing eventually. Also,
\[\begin{align*} S_{2(n + 1) + 1} = s_{2n + 1} + (b_{2n + 1} - b_{2(n + 1) + 1}) \geq s_{2n + 1} \end{align*}\]so \((s_{2m + 1})_{m = 0}^\infty\) is increasing.
We have that \(\forall n \gt m \geq N_0\), \(s_{2n} \geq s_{2m + 1}\) since
\[\begin{align*} s_{2n} = s_{2m + 1} + (b_{2n + 2} - b_{2n + 3}) + \dots + (b_{2n - 2} - b_{2n - 1}) + b_{2n} \geq s_{2m + 1} \end{align*}\]so \((s_{2n})\) is bounded below. Similarily,
\[\begin{align*} s_{2n + 1} = s_{2m} - (b_{2m + 1} - b_{2m + 2}) - \dots - (b_{2n - 1} - b_{2n}) - b_{2n + 1} \leq s_{2m} \end{align*}\]so \((s_{2n + 1})\) is bounded above. By the monotone convergence theorem, \(\lim_{m \to \infty} s_{2m} = \alpha, \lim_{m \to \infty} {s_{2m + 1}} = \beta\). For a proof of contradictino, suppose \(\alpha > \beta\). Choose \(\epsilon = \frac{\alpha - \beta}{3}\). Choose \(N_1 \geq N_0 / 2\) such that \(\vert s_{2m - \alpha} \vert \lt \epsilon, \vert s_{2m + 1} - \beta \vert \lt \epsilon\), \(\vert b_{N_1 + 1} \vert \lt \epsilon\) for all \(m \gt N_1 / 2\). Then
\(\begin{align*} \vert \alpha - \beta \vert &= \vert \alpha - s_{N_1} + s_{N_1} - s_{N_1 + 1} + s_{N_1 + 1} - \beta \\ &\leq \vert s_{N_1} - \alpha \vert + \vert s_{N_1 + 1} - s_{N_1} \vert + \vert s_{N_1 + 1} - \beta \vert \\ &\lt 3 \epsilon \\ &= \vert \alpha - \beta \vert \end{align*}\) a contradiction.
We claim that since \(\lim_{m \to \infty} s_{2m} = \alpha = \lim_{m \to \infty} s_{2m + 1}\), then \(\lim_{n \to \infty} s_n = \alpha\) (TODO). \(\blacksquare\)
Definition: A rearrangement of \(\sum a_n\) is any series \(\sum a_{\sigma(n)}\) where \(\sigma: \mathbb N \leftrightarrow \mathbb N\).
Example:
\[\begin{align*} \sum_{n = 1}^\infty \frac{(-1)^n}{n} &= -1 + 1/2 - 1/3 + 1/4 - \dots \\ &= \left( -1 - \frac{1}{3} - \frac{1}{5} - \frac{1}{7}- \frac{1}{9} - \frac{1}{11} \dots \right) + (\frac{1}{2} + \frac{1}{4}) + (- \frac{1}{13} - \dots - \frac{1}{21}) \end{align*}\]Theorem (Riemann): Let \(\sum_{n = 0}^\infty\) be conditionally convergent with \(\alpha \in \mathbb R\) fixed. Then, there exists a \(\sigma: \mathbb N \leftrightarrow \mathbb N\) such that \(\sum_{n = 0}^\infty a_{\sigma(n)} = \alpha\).
Proof: For any \(n \in \mathbb N\), \(a_n^+ = \max \lbrace 0, a_n \rbrace, a_n^- = \max \lbrace 0, -a_n \rbrace\). Then \(\forall n, a_n^+, a_n^- \geq 0, a_n = a_n^+ - a_n^-, \vert a_n \vert = a_n^+ + a_n^-\).
We show that \(\sum_{n = 0}^\infty a_n^+ = \infty, \sum_{n = 0}^\infty a_n^- = \infty\). Suppose \(\sum_{n = 0}^\infty a_n^+ \lt \infty\). Then, \(\sum_{n = 0}^\infty a_n^- = \sum_{n = 0}^\infty (a_n^+ - a_n) = \sum_0^\infty a_n^+ \sum_0^\infty a_n \lt \infty\). Then, \(\sum_0^\infty \vert a_n \vert = \sum_0^\infty (a_n^+ - a_n^-) = \sum_0^\infty a_n^+ + \sum_0^\infty a_n^- \lt \infty\); a contradiction. Similarily, if \(\sum a_n^- \lt \infty\), then one shows \(\sum a_n^+ \lt \infty\), hence \(\sum \vert a_n \vert \lt \infty\); a contradiction.
Pick \(\alpha \in \mathbb R\). Define \((b_k)_{k = 0}^\infty = (a_{n_k})_{k = 0}^\infty\) as the sequence of all the terms \(a_n\) such that \(a_n = a_n^+\). Also define \((c_l)_{l = 0}^\infty\) as the remaining terms of \((a_n)\) (i.e. such that \(a_n = a_n^- \land a_n \lt 0\)).
Define \(\sigma: \mathbb N \leftrightarrow \mathbb N\) as follows. Set \(\sigma(0) = 0\). So, \(a_\sigma(0) = a_0, s_0 = a_0\), where \((s_N)_{N = 0}^\infty\) is the sequence of partial sums of \(\sum_{n = 0}^\infty a_{\sigma(n)}\).
Then, if \(s_0 \leq \alpha\), set \(a_{\sigma(1)}, \dots, a_{\sigma(N_1)}\) to be consecutive terms of \((b_k)\) until \(s_{N_1} \gt \alpha\). Next, set \(a_{\sigma(N_1 + 1)}, \dots, a_{\sigma(N_2)}\) to be consecutive terms of \((c_l)\), until \(S_{N_2} \leq \alpha\), etc.
Otherwise, if \(s_0 \gt \alpha\), set \(a_{\sigma(1)}, \dots, a_{\sigma(N_1)}\) to be the consecutive terms of \((c_l)\) until \(S_{N_1} \leq \alpha\), etc.
We claim that \(\lim_{N \to \infty} s_N = \alpha\). Show that \(\vert s_{N_k} - \alpha \vert \leq \vert a_{\sigma(N_k)} \vert\) and \(\lim_{k \to \infty} a_{\sigma(N_k)} = 0\) because \(\lim_{n \to \infty} a_n = 0\) (TODO). \(\blacksquare\)
Theorem: Suppose \(\sum_{n = 0}^\infty a_n\) is absolutely convergent, and let \(\sigma: \mathbb N \leftrightarrow \mathbb N\) be any bijection. Then, \(\sum_{n = 0}^\infty a_{\sigma(n)}\) is absolutely convergent, and \(\sum_{n = 0}^\infty a_{\sigma(n)} = \sum_{n = 0}^\infty a_n\).
Definition: A decimal expansion of \(x \in \mathbb R\) is a sequence \((a_n)_{n = 0}^\infty\) such that
We write, \(x = a_0.a_1a_2a_3\dots\)
Proposition:
Proof:
Example: \(1.000 \dots = 0.999 \dots\) since \(\sum_{n = 1}^\infty \frac{9}{10^n} = \frac{9}{10} \sum_{n = 0}^\infty \frac{1}{10^n} = \frac{9}{10} \cdot \frac{1}{1-\frac{1}{10}} = 1\).