[Understanding Analysis by Stephen Abbott] - [Chapter 2 Sequences and Series] - [2.4 The Monotone Convergence Theorem and a First Look at Infinite Series]

Definition 2.4.1 A sequence \((a_n)\) is increasing if \(a_n \leqslant a_{n+1}\) for all \(n \in N\) and decreasing if \(a_n \geqslant a_{n+1}\) for all \(n \in N\). A sequence is monotone if it is either increasing of decreasing.

Theorem 2.4.2 Monotone Convergence Theorem. If a sequence is monotone and bounded, then it converges.

Proof. Let \((a_n)\) be monotone and bounded. To prove \((a_n)\) converges using the definition of convergence, we are going to need a candidate for the limit. Let's assume the sequence is increasing (the decreasing case is handled similarly), and consider the set of points \(\{a_n : n \in N\}\). By assumption, this set is bounded, so we can let $$ s = sup\{a_n : n \in N\} $$ It seems reasonable to claim that \(\lim a_n = s\)

To prove this, let \(\epsilon > 0\). Because \(s\) is the least upper bound for \(\{a_n : n \in N\}\), \(s - \epsilon\) is not an upper bound, so there exists a point in the sequence \(a_N\) such that \(s - \epsilon < a_N\). Now, the fact that \(a_n\) is increasing implies that if \(n \geqslant N\), then \(a_N \leqslant a_n\). Hence, $$ s - \epsilon < a_N \leqslant a_n \leqslant s < s + \epsilon $$ which implies \(|a_n - s| < \epsilon\), as desired.

The Monotone Convergence Theorem is extremely useful for the study of infinite series, largely because it asserts the convergence of a sequence without explicit mention of the actual limit.

Definition 2.4.3 Convergence of a Series. Let \((b_n)\) be a sequence. An infinite series is a formal expression of the form $$ \sum_{i=1}^\infty b_n = b_1 + b_2 + b_3 + b_4 + b_5 + \cdot \cdot \cdot $$ We define the corresponding sequence of partial sums \((s_m)\) by $$ s_m = b_1 + b_2 + b_3 + \cdot \cdot \cdot + b_m $$ and say that the series \(\sum_{n=1}^\infty b_n \) converges to \(B\) if the sequence \((s_m)\) converges to \(B\). In this case, we write \(\sum_{n=1}^\infty b_n = B\)

Theorem 2.4.6 Cauchy Condensation Test. Suppose \((b_n)\) is decreasing and satisfying \(b_n \geqslant 0\) for all \(n \in N\). Then, the series \(\sum_{n=1}^\infty b_n\) converges if and only if the series $$ \sum_{n=0}^\infty 2^n b_{2^n} = b_1 + 2b_2 + 4b_4 + 8b_8 + 16b_{16} + \cdot \cdot \cdot $$ converges.

Proof. First, assume that \(\sum_{n=0}^\infty 2^n b_{2^n}\) converges. Then the partial sums $$ t_k = b_1 + 2b_2 + 4b4 + \cdot \cdot \cdot + 2^k b_{2^k} $$ are bounded; that is, there exists an \(M > 0\) such that \(t_k \leqslant M\) for all \(k \in N\). We want to prove that \(\sum_{n=1}^\infty b_n\) converges. Because \(b_n \geqslant 0\), we know that the partial sums are increasing, so we only need to show that $$ s_m = b_1 + b_2 + b_3 \cdot \cdot \cdot + b_m $$ is bounded.

Fix \(m\) and let \(k\) be large enough to ensure \(m \leqslant 2^{k+1} - 1\). Then, \(s_m \leqslant s_{2^{k+1} - 1}\) and $$ \begin{align} s_{2^{k+1} - 1} &= b_1 + (b_2 + b_3) + (b_4 + b_5 + b_6 + b_7) + \cdot \cdot \cdot + (b_{2^k} + \cdot \cdot \cdot + b_{2^{k+1} - 1}) &\leqslant b_1 + (b_2 + b_2) + (b_4 + b_4 + b_4 + b_4) + \cdot \cdot \cdot + (b_{2^k} + \cdot \cdot \cdot + b_{2^k} ) &= b_1 + 2b_2 + 4b_4 + \cdot \cdot \cdot + 2^k b_{2^k} = t_k \end{align} $$ Thus, \(s_m \leqslant t_k \leqslant M\). By the Monotone Convergence Theorem, we can conclude that \(\sum_{n=1}^\infty b_n\) converges.

Corollary 2.4.7 The series \(\sum_{n=1}^\infty 1/n^p\) converges if and only if \(p > 1\)