Exercises §14 — Ross, Elementary Analysis, 2nd Edition
Exercise 14.9. The convergence of a series does not depend on any finite number of the terms, but the value of the limit does. Consider series \(\sum a_n\) and \(\sum b_n\) and suppose \(\{\, n \in \mathbb{N} : a_n \neq b_n \,\}\) is finite. Show that this means the series both converge or both diverge.
Proof. Since \(\{\, n \in \mathbb{N} : a_n \neq b_n \,\}\) is finite, choose \(M \in \mathbb{N}\) such that \(a_n = b_n\) for all \(n > M\). A series converges if and only if it satisfies the Cauchy criterion, that is, for all \(\varepsilon > 0\) there exists \(N_0\) such that \[ n \ge m \ge N_0 \implies \left| \sum_{k=m}^{n} a_k \right| < \varepsilon. \] Let \(N = \max(M, N_0)\). Then \[ n \ge m \ge N \implies \left| \sum_{k=m}^{n} a_k \right| = \left| \sum_{k=m}^{n} b_k \right|, \] so either they both converge, or they both diverge if they cannot satisfy the Cauchy criterion. \(\square\)
Exercise 14.12. Let \((a_n)_{n \in \mathbb{N}}\) be a sequence such that \(\liminf |a_n| = 0\). Prove there is a subsequence \((a_{n_k})_{k \in \mathbb{N}}\) such that \(\sum_{k=1}^{\infty} a_{n_k}\) converges.
Proof. Since \(\liminf |a_n| = 0\), that means for all \(\varepsilon > 0\) and all \(N \in \mathbb{N}\) there exists \(n \ge N\) such that \(|a_n| < \varepsilon\). So let’s create a subsequence \((a_{n_k})\) such that \[ |a_{n_k}| < 2^{-k}, \quad k \in \mathbb{N}. \]
Note \[ 0 \le \sum_{k=1}^{m} |a_{n_k}| \le \sum_{k=1}^{m} \frac{1}{2^k} \le 1. \] Since the partial sums are increasing and bounded above, the series \(\sum_{k=1}^{\infty} a_{n_k}\) converges absolutely, which means \[ \sum_{k=1}^{\infty} a_{n_k} \text{ converges.} \quad \square \]