Theorem 28.1
Theorem 28.1 (Summation by Parts). Let \( \sum_{n=1}^{\infty} a_n \) be an infinite series and let \( \{s_n\} \) be the sequence of partial sums of \( \sum_{n=1}^{\infty} a_n \). Let \( \{b_n\} \) be any sequence. Then for any positive integer \( n \), we have \[ \sum_{k=1}^{n} a_k b_k = \sum_{k=1}^{n} s_k (b_k - b_{k+1}) + s_n b_{n+1}. \]
Proof. Let $s_0 = 0$. Then $$\sum_{k=1}^n a_kb_k = \sum_{k=1}^n (s_k - s_{k-1})b_k \\ = \sum_{k=1}^n s_kb_k - \sum_{k=1}^n s_{k-1}b_k \\ = \sum_{k=1}^n s_kb_k - \sum_{k=1}^n s_kb_{k+1} - s_nb_{n+1} \\ = \sum_{k=1}^n s_k(b_k - b_{k+1}) - s_nb_{n+1}. \blacksquare$$
Theorem 28.2
Theorem 28.2. Let \( \sum_{n=1}^{\infty} a_n \) be a series whose sequence of partial sums is bounded. If \( \sum_{n=1}^{\infty} |b_n - b_{n+1}| \) converges and \( \lim_{n \to \infty} b_n = 0 \), then \( \sum_{n=1}^{\infty} a_n b_n \) converges.
Proof. Let $\{ s_n \}$ be the sequence of partial sums of the series $\sum_{n=1}^{\infty} a_n$. Since $\sum_{n=1}^{\infty} (b_n - b_{n+1})$ converges absolutely and $\{ s_n \}$ is bounded, \sum_{n=1}^{\infty} s_n(b_n - b_{n+1}) converges absolutely by Theorem 26.4.
Since $\lim_{n \to \infty} b_n = 0$ and $\{ s_n \}$ is bounded, $\lim_{n \to \infty} s_nb_{n+1} = 0$ by Theorem 13.3 and Theorem 11.2.
Thus $$\sum_{n=1}^{\infty} a_nb_n = \sum_{n=1}^{\infty} s_n(b_n - b_{n+1}) + \lim_{n \to \infty} s_nb_{n+1}$$ converges by Theorem 28.1. $\blacksquare$
Dirichlet's Test
Corollary 28.3 (Dirichlet's Test). If the sequence of partial sums of the series \( \sum_{n=1}^{\infty} a_n \) is bounded and \( \{b_n\} \) is a decreasing sequence with limit 0, then \( \sum_{n=1}^{\infty} a_n b_n \) converges.
Proof. The sequence of partial sums of $\sum_{n=1}^{\infty} |b_n - b_{n+1}|$ is given by $$s_n = |b_1 - b_2| + \cdots + |b_n - b_{n+1}| \\ = (b_1 - b_2) + \cdots + (b_n - b_{n+1}) \quad (\because) \{ b_n \} \text{ is decreasing} \\ = b_1 - b_{n+1}.$$ Since $\{ s_n \}$ converges, $\sum_{n=1}^{\infty} a_nb_n$ converges by Theorem 28.2. $\blacksquare$
Corollary 28.4
Corollary 28.4 (Alternating Series Test Revisited). If \( \{b_n\} \) is a decreasing sequence with limit 0, then \[ \sum_{n=1}^{\infty} (-1)^{n+1} b_n \] converges.
Proof. Since the sequence of partial sums of $\sum_{n=1}^{\infty} (-1)^{n+1}$ is bounded, $\sum_{n=1}^{\infty} (-1)^{n+1} b_n$ converges by Dirichlet's Test. $\blacksquare$
Theorem 28.5
Theorem 28.5. If \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} |b_n - b_{n+1}| \) are convergent series, then \( \sum_{n=1}^{\infty} a_n b_n \) converges.
Proof. Note that the sequence $\{ s_n \}$ of partial sums of $\sum_{n=1}^{\infty} a_n$ is bounded. Then $\sum_{n=1}^{\infty} s_n(b_n - b_{n+1})$ converges absolutely by Theorem 26.4.
Since $\sum_{n=1}^{\infty} (b_n - b_{n+1})$ converges by Theorem 26.2, $$\lim_{n \to \infty} ((b_1 - b_2) + \cdots + (b_n - b_{n+1})) \\ = \lim_{n \to \infty} (b_1 - b_{n+1})$$ converges, so that $\lim_{n \to \infty} b_{n+1}$ converges. Thus $\{ s_nb_{n+1} \}$ converges by Theorem 12.6. By Theorem 28.1, $\sum_{n=1}^{\infty} a_nb_n$ converges. $\blacksquare$
Abel's Test
Corollary 28.6 (Abel's Test). If \( \sum_{n=1}^{\infty} a_n \) converges and \( \{b_n\} \) is a bounded monotone sequence, then \( \sum_{n=1}^{\infty} a_n b_n \) converges.
Proof. If $\{ b_n \}$ is increasing, then $$\sum_{n=1}^{k} |b_n - b_{n+1}| = |b_1 - b_2| + \cdots + |b_k - b_{k+1}| = -(b_1 - b_2) - \cdots - (b_k - b_{k+1}) \\ = -b_1 + b_{k+1}.$$ Since $\{ b_n \}$ converges by Theorem 16.2, $\sum_{n=1}^{\infty} |b_n - b_{n+1}|$ converges. Similarly, we can show that if $\{ b_n \}$ is decreasing, $\sum_{n=1}^{\infty} |b_n - b_{n+1}|$ converges. Thus the conclusion follows immediately from Theorem 28.5. $\blacksquare$