Prove Theorems 2.1, 2.2, and 2.3.
Solution
1. Theorem 2.1 (Monotone Convergence Theorem for series).Proof: let \(a_n \ge 0\).The partial sums \(S_N = \sum_{n=1}^N a_n\) form an increasing sequence, so \(S_N\) converges in \([0,\infty]\); the series converges iff \((S_N)\) is bounded above, and then the sum is \(\sup_N S_N\).(If \(a_n \ge 0\) only eventually, ignore the initial terms.)
2. Theorem 2.2 (nth-term test).Proof: if \(\sum a_n\) converges with partial sums \(S_N \to S\), then \(a_n = S_n - S_{n-1} \to S - S = 0\).Hence if \(a_n \not\to 0\) the series diverges.(For series of real numbers; for a series of vectors the same holds via Cauchy: \(a_n \to 0\) is necessary.)
3. Theorem 2.3 (Cauchy condensation test).Proof: assume \(a_1 \ge a_2 \ge \cdots \ge 0\).(⇒) If \(\sum a_n\) converges, group terms between powers of \(2\): \[\sum_{n=1}^{2^m - 1} a_n \ge a_1 + (a_2 + a_3) + \cdots + (a_{2^{m-1}} + \cdots + a_{2^m - 1}) \ge a_1 + \frac12\big(2a_2 + 4a_4 + \cdots + 2^{m-1}a_{2^{m-1}}\big),\] so the partial sums of \(\sum 2^k a_{2^k}\) are bounded by \(2\sum a_n < \infty\); hence \(\sum 2^k a_{2^k}\) converges.(⇐) If \(\sum 2^k a_{2^k}\) converges, then grouping in the other direction: \[\sum_{n=2}^{2^m} a_n \le (a_2 + a_3) + (a_4 + \cdots + a_7) + \cdots + (a_{2^{m-1}} + \cdots + a_{2^m}) \le 2a_2 + 4a_4 + \cdots + 2^m a_{2^m},\] so the partial sums of \(\sum a_n\) are bounded, hence \(\sum a_n\) converges.
4. Q.E.D.