Seperated Sequences
Given a sequence \(a_n (n \in \mathbb{N})\), we can seperate all terms into two sequences:
\[a_{n_1}, a_{n_2}, \dots \text{ and } a_{n_1'}, a_{n_2'}, \dots\]with
\[n_1 < n_2 < \dots \text{ and } n_1' < n_2' < \dots\]also
\[\begin{align} \{n_1, n_2, \dots\} \cup \{n_1', n_2', \dots\} &= \mathbb{N}\\ \{n_1, n_2, \dots\} \cap \{n_1', n_2', \dots\} &= \emptyset \end{align}\]s.t. \(a_{n_j} \ge 0 (j \in \mathbb{N})\) and \(a_{n_k'} \le 0 (k \in \mathbb{N})\)
Let \(p_j := a_{n_j} (j \in \mathbb{N})\) and \(q_k := - a_{n_k'} (k \in \mathbb{N})\), we know \(p_j \ge 0, q_k \ge 0\).
Exercise 3: Show that
\[\sum_{n} \vert a_n \vert < \infty \iff \sum_{j} p_j < \infty \land \sum_{k} q_k < \infty\]Moreover, if and side holds, then
\[\sum_{n} \vert a_n \vert = \sum_{j} p_j + \sum_k q_k\] \[\sum_{n} a_n = \sum_{j} p_j - \sum_k q_k\]Proof:
\(\implies\): Let the partial sum of \(\sum_n \vert a_n \vert\) has upper bound M.
The partial sum of \(\sum_j p_j\) is less or equal to the partial sum of \(\sum_n \vert a_n \vert\)
\[\begin{align} &p_1 + \dots + p_k = a_{n_1} + \dots + a_{n_k}\\ \le & \vert a_1 \vert + \vert a_2 \vert + \dots + \vert a_{n_k} \vert \le M \end{align}\]Thus the right side has upper bound, thus converges.
\(\Longleftarrow\): Let the partial sum of \(\sum_j p_j, \sum_k q_k\) has uppber bound \(U, V\) respectively. Consider the partial sum of \(\sum_n \vert a_n \vert\)
\[\vert a_1 \vert + \dots + \vert a_k \vert \le p_1 + \dots + p_k + q_1 + \dots + q_k \le U + V\]To prove \(\sum_{n} \vert a_n \vert = \sum_{j} p_j + \sum_k q_k\), let \(s_k = \vert a_1 \vert + \dots + \vert a_k \vert\), \(u_k = p_1 + \dots + p_k\), \(v_k = q_1 + \dots + q_k\), and let
\[\lim_{k\to \infty} u_k = L_p\] \[\lim_{k\to\infty} v_k = L_q\]We know
\[s_k \le u_k + v_k \implies \lim_{k \to \infty} s_k \le L_p + L_q\]also
\[u_k + v_k \le s_{\max(n_k, n'_k)}\]It is easy to see \(\lim_{k\to\infty} s_{\max(n_k,n'_k)} = \lim_{k\to\infty} s_k\), thus
\[\lim_{k\to\infty} s_k \ge L_p + L_q\]Thus \(\sum_{n} \vert a_n \vert = \sum_{j} p_j + \sum_k q_k\).
To prove \(\sum_{n} a_n = \sum_j p_j - \sum_k q_k\), now let \(s_k = a_1 + \dots + a_k\)
\[\forall \varepsilon > 0 \exists N \in \mathbb{N} [k \ge N \implies \vert u_k - L_p \vert < \varepsilon \land \vert v_k - L_q \vert < \varepsilon]\] \[k \ge \max(n_N, n_N') \implies s_{k} = u_{N_1} - v_{N_2}, \text{ where } N_1 \ge N, n_2 \ge N\] \[k \ge \max(n_N, n_N') \implies \vert s_{k} - (L_p - L_q) \vert = \vert (u_{N_1} - L_p) + (v_{N_2} - L_q) \vert \le 2 \varepsilon\]Thus \(\sum_{n} a_n = \sum_j p_j - \sum_k q_k\).
Two Convergent Series
- \(\sum_n a_n\) and \(\sum_n b_n\) converge \(\implies \sum_{n} (a_n \pm b_n) = \sum_n a_n \pm \sum_n b_n\)
- Inserting finite number of 0s into all the two neighbor terms, will not affect its convergence or divergence and its sum (if the sum exists).
16min