Theorem 24.1
Theorem 24.1. Let $\sum_{n=1}^{\infty} a_n$ be a series with nonnegative terms. Then $\sum_{n=1}^{\infty} a_n$ converges $\iff$ the sequence of partial sums $\{ s_n \}$ is bounded.
Proof. $(\Longrightarrow)$
Since $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} s_n$ converges, $\{ s_n \}$ is bounded by Theorem 13.2.
$(\Longleftarrow)$
Since $a_n \geq 0, \forall n \in \mathbb{P}$, $\{ s_n \}$ is monotonic increasing. By Theorem 16.2, $\lim_{n \to \infty} s_n = \sum_{n=1}^{\infty} a_n$ converges. $\blacksquare$
Remark
Remark. If a series $\sum_{n=1}^{\infty} a_n$ with nonnegative terms converges to $L$ and $\{ s_n \}$ is the sequence of partial sums of $\sum_{n=1}^{\infty} a_n$, then $s_n \leq L, \forall n \in \mathbb{P}$. It may also be written $\sum_{k=1}^n a_k \leq \sum_{k=1}^{\infty} a_k$.
$(\because)$
Since $\{ s_n \}$ converges, it is bounded. Then the set $S = \{ s_n \, | \, n \in \mathbb{P} \}$ has the least upper bound $M$. For any $\varepsilon > 0$, $M - \varepsilon$ is not an upper bound for $S$. Then $s_N > M - \varepsilon$ for some positive integer $N$. Since $\{ s_n \}$ is increasing, $M - \varepsilon < M, \forall n \geq N$. This means that $\lim_{n \to \infty} s_n = M = L$. Thus $L$ is the least upper bound for $S$, which means that $s_n \leq L, \forall n \in \mathbb{P}$.
$2^n$ Test
Theorem 24.2. ($2^n$ Test) Let \(\{a_n\}\) be a decreasing sequence of nonnegative numbers. Then \(\sum_{n=1}^{\infty} a_n\) converges if and only if \(\sum_{n=1}^{\infty} 2^n a_{2^n}\) converges.
Proof. $(\Longleftarrow)$
Suppose that $\sum_{n=1}^{\infty} 2^n a_{2^n}$ converges. Note that $$a_1 \leq a_1 \\ a_2 + a_3 \leq 2a_2 \\ a_4 + a_5 + a_6 + a_7 \leq 4a_4 \\ \vdots \\ a_{2^n} + \cdots + a_{2^{n+1}-1} \leq 2^n a_{2^n} \\ \Longrightarrow \sum_{k=1}^{2^{n+1}-1} a_k \leq \sum_{k=1}^{n} 2^k a_{2^k} \leq \sum_{k=1}^{\infty} 2^k a_{2^k}.$$ Thus partial sum $\{ s_n \}$ of the series is bounded. Since $\{ s_n \}$ is increasing, $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} s_n$ converges by Theorem 16.2.
$(\Longrightarrow)$
Assume that $\sum_{n=1}^{\infty} a_n$ converges. Note that $$a_3 + a_4 \geq 2 a_4 \\ a_5 + a_6 + a_7 + a_8 \geq 4a_8 \\ \vdots \\ a_{2^n + 1} + \cdots + a_{2^{n+1}} \geq 2^n a_{2^{n+1}} \\ \Longrightarrow \sum_{k=1}^{\infty} a_k \geq \sum_{k=3}^{2^{n+1}} a_k \geq \sum_{k=3}^n 2^k a_{2^{k+1}} = \frac{1}{2} \sum_{k=1}^n 2^{k+1} a_{2^{k+1}} \\ \Longrightarrow \sum_{k=1}^n 2^k a_{2^k} = 2a_2 + \sum_{k=1}^n 2^{k+1} a_{2^{k+1}} \leq 2\sum_{k=3}^{\infty} a_k + 2a_2.$$ Thus $t_n = \sum_{k=1}^n 2^k a_{2^k}$ is bounded. Since $\{ t_n \}$ is increasing, it is convergent by Theorem 16.2. $\blacksquare$
P-Series Test
Corollary 24.3. (P-Series Test) The series \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) diverges if \(p < 1\) and converges if \(p > 1\).
Proof. Note that $$\sum_{n=1}^{\infty} 2^n \frac{1}{(2^{n})^p} = \sum_{n=1}^{\infty} 2^n 2^{-np} = \sum_{n=1}^{\infty} 2^{n(1-p)} = \sum_{n=1}^{\infty} (2^{1-p})^n.$$ If $1-p \geq 0$, then $2^{1-p} \geq 2^0 = 1$ and if $1-p < 0$, then $2^{1-p} < 2^0 = 1$. Since $\sum_{n=1}^{\infty} (2^{1-p})^n$ is a geometric series, $\sum_{n=1}^{\infty} n^{-p}$ diverges if $p \leq 1$ and converges if $p > 1$ by $2^n$ test. $\blacksquare$
Alternating Series Test
Theorem 25.1. (Alternating Series Test) Let \(\{a_n\}\) be a decreasing sequence such that \(\lim_{n \to \infty} a_n = 0\). Then \(\sum_{n=1}^{\infty} (-1)^{n+1} a_n\) converges.
Proof. Note that $a_n \geq 0, \forall n \in \mathbb{P}$. ($(\because)$ If $a_N < 0$ for some positive integer $N$, then $a_n < 0, \forall n \geq N$ because $\{ a _n \}$ is decreasing. Thus $\lim_{n \to \infty} a_n = - \infty$ by Definition 15.1. $\bigotimes$)
Let $\{ s_n \}$ be partial sums of $\sum_{n=1}^{\infty} a_n$. Since $\{ a_n \}$ is decreasing, $a_{2n+1} \geq a_{2n+2}, \forall n \in \mathbb{P}$. Then we have $s_{2(n+1)} - s_{2n} = -a_{2n+2} + a_{2n+1} \geq 0, \forall n \in \mathbb{P}$, which means that $\{ s_{2n} \}$ is increasing.
Note that $$s_{2n} = a_1 - a_2 + \cdots + a_{2n-1} - a_{2n} \\ = a_1 - (a_2 - a_3) - \cdots - (a_{2n-2} - a_{2n-1}) - a_{2n} \leq a_1 \\ (\because) a_n - a_{n+1} \geq 0, \forall n \in \mathbb{P}.$$ Thus $\{ s_{2n} \}$ is bounded. Then $\{ s_{2n} \}$ converges by Theorem 16.2.
Since $s_{2n - 1} = s_{2n} + a_{2n}$, we have $$\lim_{n \to \infty} s_{2n - 1} = \lim_{n \to \infty} (s_{2n} + a_{2n}) = \lim_{n \to \infty} s_{2n} + \lim_{n \to \infty} a_{2n} = \lim_{n \to \infty} s_{2n}.$$ Since $\{ s_{2n} \}$ and $\{ s_{2n-1} \}$ converges to the same limit, $\{ s_n \}$ converges. $\blacksquare$
Corollary 25.2
Corollary 25.2. If \(s > 0\), then \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s}\) converges.
Proof. Note that for any positive integer $n$, $$n < n+1 \Longrightarrow n^{-s} > (n+1)^{-s}.$$ Thus the sequence $\{ n^{-s} \}$ is decreasing and $\lim_{n \to \infty} n^{-s} = 0$.
($(\because)$ Let any $\varepsilon > 0$. Then $\exists N \in \mathbb{P}$ such that $\frac{1}{N} < \varepsilon^{\frac{1}{s}}$ by Corollary 6.12. Note that $n^{-s} > 0, \forall n \in \mathbb{P}$ because if not, then $N_0^{-s} \leq 0$ for some positive integer $N_0$, so that $$n^{-s} \leq 0, \forall n \geq N_0 \\ \Longrightarrow 1 \leq 0 \bigotimes.$$ Thus we have that for $n \geq N$, $$|n^{-s}| = n^{-s} \leq N^{-s} < \varepsilon.$$ Thus $\lim_{n \to \infty} n^{-s} = 0$.)
By Theorem 25.1, $\sum_{n=1}^{\infty} (-1)^{n+1} n^{-s}$ converges. $\blacksquare$
Absolute Convergence
Definition 26.1. Let \(\sum_{n=1}^{\infty} a_n\) be an infinite series. If \(\sum_{n=1}^{\infty} |a_n|\) converges, we say \(\sum_{n=1}^{\infty} a_n\) converges absolutely. If \(\sum_{n=1}^{\infty} a_n\) converges and \(\sum_{n=1}^{\infty} |a_n|\) diverges, we say \(\sum_{n=1}^{\infty} a_n\) converges conditionally.
Theorem 26.2
Theorem 26.2. If \(\sum_{n=1}^{\infty} a_n\) converges absolutely, then \(\sum_{n=1}^{\infty} a_n\) converges and \(\left| \sum_{n=1}^{\infty} a_n \right| \leq \sum_{n=1}^{\infty} |a_n|\).
Proof. Let $$p_n = \begin{cases} a_n & \text{if } a_n \geq 0 \\ 0 & \text{if } a_n < 0 \end{cases}$$ and $$q_n = \begin{cases} 0 & \text{if } a_n \geq 0 \\ -a_n & \text{if } a_n < 0. \end{cases}$$ Then we have $a_n = p_n - q_n$ and $|a_n| = p_n + q_n$ for all positive integers $n$.
( $(\because)$ If $a_n \geq 0$, then $p_n = a_n$ and $q_n = 0$, so that $a_n = a_n + 0 = p_n - q_n$ and $|a_n| = a_n + 0 = p_n + q_n$. If $a_n < 0$, then $p_n = 0$ and $q_n = -a_n$, so that $a_n = 0 - (-a_n) = p_n - q_n$ and $|a_n| = 0 + (-a_n) = p_n + q_n$.)
Let $\{ s_n \}$ and $\{ t_n \}$ be two partial sums of the series $\sum_{n=1}^{\infty} p_n$ and $\sum_{n=1}^{\infty} q_n$, respectively. Note that $$s_n = p_1 + \cdots + p_n \leq |a_1| + \cdots + |a_n| = \sum_{k=1}^n |a_k|, \forall n \in \mathbb{P},$$ because if $a_n \geq 0$, then $p_n = a_n \Longrightarrow p_n \leq |a_n|$ and if $a_n < 0$, then $p_n = 0 \Longrightarrow p_n \leq |a_n|$, which means that $p_n \leq |a_n|, \forall n \in \mathbb{P}.$ Since $\sum_{n=1}^{\infty} a_n$ converges absolutely, $s_n \leq \sum_{n=1}^{\infty} |a_n|, \forall n \in \mathbb{P}$ by Remark. Thus $\{ s_n \}$ is bounded.
Since all $n$th terms of $\{ p_n \}$ is nonnegative, $\{ s_n \}$ is increasing. By Theorem 16.2, $\sum_{n=1}^{\infty} p_n$ converges.
Similarly, we have that $$t_n = q_1 + \cdots + q_n \leq |a_1| + \cdots + |a_n| = \sum_{k=1}^{n} |a_k|, \forall n \in \mathbb{P},$$ because if $a_n \geq 0$, then $q_n = 0 \Longrightarrow q_n \leq |a_n|$ and if $a_n < 0$, then $q_n = -a_n \Longrightarrow q_n \leq |a_n|$, which means that $q_n \leq |a_n|, \forall n \in \mathbb{P}.$ Since $\sum_{n=1}^{\infty} a_n$ converges absolutely, $t_n \leq \sum_{n=1}^{\infty} |a_n|, \forall n \in \mathbb{P}$ by Remark. Thus $\{ t_n \}$ is bounded.
Since all $n$th terms of $\{ t_n \}$ is nonnegative, $\{ t_n \}$ is increasing. By Theorem 16.2, $\sum_{n=1}^{\infty} q_n$ converges.
Thus $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (p_n - q_n) = \sum_{n=1}^{\infty} p_n - \sum_{n=1}^{\infty} q_n$ converges by Theorem 23.1.
Note that $|\sum_{k=1}^n a_k| = |a_1 + \cdots + a_n| \leq |a_1| + \cdots + |a_n| = \sum_{k=1}^n |a_k|, \forall n \in \mathbb{P}$ by the triangle inequality. Taking the limits, we have that $|\sum_{n=1}^{\infty} a_n | \leq \sum_{n=1}^{\infty} |a_n|$ by Exercise 10.12. $\blacksquare$
Comparison Test
Theorem 26.3 (Comparison Test) Let \(\sum_{n=1}^{\infty} a_n\) and \(\sum_{n=1}^{\infty} b_n\) be two series such that \[ |a_n| \leq |b_n| \] for every positive integer \(n\).
(i) If \(\sum_{n=1}^{\infty} b_n\) converges absolutely, then \(\sum_{n=1}^{\infty} a_n\) converges absolutely and \[ \sum_{n=1}^{\infty} |a_n| \leq \sum_{n=1}^{\infty} |b_n|. \]
(ii) If \(\sum_{n=1}^{\infty} |a_n|\) diverges, then \(\sum_{n=1}^{\infty} a_n\) diverges.
Proof. (i) Note that $\sum_{k=1}^n |a_k| \leq \sum_{k=1}^n |b_k| \leq \sum_{k=1}^{\infty} |b_k|, \forall n \in \mathbb{P}$, so that the sequence of partial sums of the series $\sum_{k=1}^{\infty} |a_k|$ is bounded. Since the sequence of partial sums of $\sum_{k=1}^{\infty} |a_k|$ is increasing, it is convergent by Theorem 16.2. Thus $\sum_{n=1}^{\infty} a_n$ converges absolutely and $$\sum_{n=1}^{\infty} |a_n| \leq \sum_{n=1}^{\infty} |b_n|$$ by Theorem 14.2.
(ii) Note that if $\sum_{n=1}^{\infty} a_n$ does not converge absolutely, then $\sum_{n=1}^{\infty} |a_n|$ does not converge, which means that it diverges. Thus (ii) is the contrapositive of (i). $\blacksquare$
Theorem 26.4
Theorem 26.4. Let \(\sum_{n=1}^{\infty} a_n\) be an infinite series and \(\{b_n\}\) be a sequence.
(i) If the sequence \(\{b_n\}\) is bounded and \(\sum_{n=1}^{\infty} a_n\) converges absolutely, then \(\sum_{n=1}^{\infty} a_n b_n\) converges absolutely.
(ii) If the sequence \(\left\{ \frac{1}{b_n} \right\}\) is bounded and \(\sum_{n=1}^{\infty} |a_n|\) diverges, then \(\sum_{n=1}^{\infty} |a_n b_n|\) diverges.
Proof. (i) Since $\{ b_n \}$ is bounded, $\exists M \in \mathbb{R}$ such that $|b_n| \leq M, \forall n \in \mathbb{P}$. Then $|a_nb_n| \leq |a_n|M, \forall n \in \mathbb{P}$. Since $\sum_{n=1}^{\infty} M a_n$ converges absolutely, $\sum_{n=1}^{\infty} a_n b_n$ converges absolutely by Comparision test.
(ii) Since $\{ \frac{1}{b_n} \}$ is bounded, $\exists M \in \mathbb{R}$ such that $$\Bigg|\frac{1}{b_n} \Bigg| \leq M, \forall n \in \mathbb{P} \\ \Longrightarrow \frac{1}{M} \leq |b_n|, \forall n \in \mathbb{P}.$$ Then $\frac{|a_n|}{M} \leq |a_nb_n|, \forall n \in \mathbb{P}$. Since $\sum_{n=1}^{\infty} \frac{|a_n|}{M}$ diverges, $\sum_{n=1}^{\infty} |a_nb_n|$ diverges by Comparison test. $\blacksquare$
Theorem 13.3과 비슷하게 이해할 수 있다. $\sum_{n=1}^{\infty} a_n$이 절대수렴하므로 수열 $\{ |a_n| \}$은 $n$이 커질수록 0으로 수렴한다. 이런 $|a_n|$에 적당한 범위 내에서 존재하는 $b_n$을 아무리 곱해봤자 그 곱인 $a_nb_n$은 여전히 0으로 수렴할 것이고, $\sum_{n=1}^{\infty} a_nb_n$ 또한 절대수렴할 것이다.
Corollary 26.5
Corollary 26.5. Let \(\sum_{n=1}^{\infty} a_n\) be a series and \(\{b_n\}\) be a convergent sequence.
(i) If \(\sum_{n=1}^{\infty} a_n\) converges absolutely, then \(\sum_{n=1}^{\infty} a_n b_n\) converges absolutely.
(ii) If \(\sum_{n=1}^{\infty} |a_n|\) diverges and \(\lim_{n \to \infty} b_n \neq 0\), then \(\sum_{n=1}^{\infty} |a_n b_n|\) diverges.
Proof. (i) Since $\{ b_n \}$ converges, it is bounded by Theorem 13.2. Then the statement follows immediately from Theorem 26.4.
(ii) Since $\lim_{n \to \infty} b_n \neq 0$, $\lim_{n \to \infty} \frac{1}{b_n}$ exists. Then $b_n \neq 0$ for all but finitely many positive integers $n$ by Lemma 12.8, so that $\{ \frac{1}{b_n} \}$ is bounded by Theorem 13.2. Thus, the statement follows from Theorem 26.4. $\blacksquare$
Ratio Test
Theorem 26.6 (Ratio Test). Let \(\sum_{n=1}^{\infty} a_n\) be a series such that \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \text{ exists } \quad (L = \infty \text{ is allowed}). \] (i) If \(L < 1\), then \(\sum_{n=1}^{\infty} a_n\) converges absolutely.
(ii) If \(L > 1\), then \(\sum_{n=1}^{\infty} a_n\) diverges.
Proof. (i) Suppose that $L < 1$. Choose $M \in \mathbb{R}$ such that $L < M < 1$. Then $M - L > 0$. Since $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$ exists, we can choose $\varepsilon$ such that $\varepsilon < M - L$. Then $\exists N$ such that $\left| \left| \frac{a_{n+1}}{a_n} \right| -L \right| < \varepsilon, \forall n \geq N$. Thus we have $\left| \frac{a_{n+1}}{a_n} \right| < L + \varepsilon < L + M - L = M, \forall n \geq N$.
Note that $$\left| \frac{a_{N+1}}{a_N} \right| < M \\ \Longrightarrow |a_{N+1}| < M|a_N|$$ and $$\left| \frac{a_{N+2}}{a_{N+1}} \right| < M \\ \Longrightarrow |a_{N+2}| < M|a_{N+1}| < M^2 |a_N|.$$ In general, we have that $|a_{N+n}| < M^n |a_N|, \forall n \in \mathbb{P}$. Note that $\sum_{n=1}^{\infty} M^n |a_N|$ is a geometric series and it converges when $0 < M < 1$. Since we choose $M$ such that $M < 1$, it converges and $\sum_{n=1}^{\infty} |a_{N+n}|$ converges by Comparison test. Since $\sum_{n=1}^k |a_{N+n}| = \sum_{n=N+1}^k |a_n| = \sum_{n=1}^k |a_n| - \sum_{n=1}^N |a_n|$ and it converges, $\sum_{n=1}^k |a_n|$ also converges. (If not, $\sum_{n=1}^k |a_{N+n}|$ diverges. $\bigotimes$) Thus $\sum_{n=1}^{\infty} a_n$ converges absolutely.
(ii) Suppose that $L > 1$. Since $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$ exists, we can choose $\varepsilon$ such that $\varepsilon < L - 1$. Then $\exists N$ such that $\left| \left| \frac{a_{n+1}}{a_n} \right| -L \right| < \varepsilon, \forall n \geq N$. Thus we have $\left| \frac{a_{n+1}}{a_n} \right| > L - \varepsilon > L - L + 1 = 1 \forall n \geq N$.
Note that $|a_N| < |a_{N+1}| < |a_{N+2}| < \cdots$ so that $\{ a_n \}$ does not converge to $0$. By Theorem 22.3, $\sum_{n=1}^{\infty} a_n$ diverges. $\blacksquare$
(($\because$) If not, $\forall \varepsilon > 0, \exists N_0$ such that $|a_n - 0| < \varepsilon, \forall n \geq N_0$. Let $N' = \max \{ N, N_0 \}$. Then $|a_n| < \varepsilon, \forall n \geq N'$, which means that )
Root Test
Theorem 26.7 (Root Test). Let \(\{a_n\}\) be a sequence and let \[ L = \limsup_{n \to \infty} |a_n|^{1/n} \quad (L = \infty \text{ is allowed}). \] (i) If \(L < 1\), then \(\sum_{n=1}^{\infty} a_n\) converges absolutely.
(ii) If \(L > 1\), then \(\sum_{n=1}^{\infty} a_n\) diverges.
Proof. (i) Suppose that $L < 1$. Choose a real number $M$ such that $L < M < 1$. By Theorem 20.3, $\exists N$ such that $|a_n|^{\frac{1}{n}} < M, \forall n \geq N$. Then $|a_n| < M^n, \forall n \geq N$. Since $\sum_{n=1}^{\infty} M^n$ converges, $\sum_{n=1}^{\infty} |a_n|$ converges by Comparision test.
(ii) Suppose that $L > 1$. Then $1 < |a_n|^{\frac{1}{n}}$ for infinitely many positive integers $n$. Thus $1 < |a_n|$ for infinitely many positive integers $n$, which means that $\{ a_n \}$ does not converge to $0$. (($\because$) If not, $\lim_{n \to \infty} |a_n| = 0$ and we have $1 \leq 0$ by Theorem 14.2. $\bigotimes$) By Theorem 22.3, $\sum_{n=1}^{\infty} a_n$ diverges. $\blacksquare$