Dirichlet Test and Absolute Convergence

Theorem 7.2.1

Let $\{a_k\}$ and $\{b_k\}$ be a sequence of real numbers. Set $A_0:=0, A_n:=\sum_{k=1}^n a_k$ for $n\geq 1$. Then if $1\leq p \leq q$,

\[\begin{align*} \sum_{k=p}^q a_k b_k = \sum_{k=p}^{q-1}A_k(b_k-b_{k+1}) + A_q b_q - A_{p-1}b_p \end{align*}\]

<Proof>

\[\begin{align*} \sum_{n=p}^q a_nb_n &= \sum_{n=p}^q(A_n - A_{n-1})b_n \\ &=\sum_{n=p}^q A_n b_n - \sum_{n=p}^q A_{n-1}b_n\\ &=\sum_{n=p}^q A_n b_n - \sum_{n=p-1}^{q-1} A_{n}b_{n+1}\\ &=\sum_{n=p}^{q-1} A_n (b_n-b_{n+1}) + A_qb_q - A_{p-1}b_p \end{align*}\] \[\tag*{$\square$}\]

Theorem 7.2.2 (Dirichlet Test)

Suppose $\{ a_k\}$ and $\{b_k\}$ are sequences of real numbers satisfying the following:

(a) the partial sums $A_n=\sum_{k=1}^n a_k$ form a bounded sequence.

(b) $b_1\geq b_2 \geq b_3 \geq \cdots \geq 0$, and

(c) $\lim_{k\to\infty}b_k = 0$

Then $\sum_{k=1}^\infty a_k b_k$ converges.

<Proof>

Since $\{A_k\}$ is bounded, there is $M>0$ such that $\lvert A_n \rvert \leq M$ for all $n\in\mathbb{N}$. Let $\epsilon >0$ be given. Since $b_n$ is bounded below and monotone decreasing, there is $N\in\mathbb{N}$ such that

\[\begin{align*} n\geq N \Rightarrow b_n < \frac{\epsilon}{2M}. \end{align*}\]

If $q\geq p \geq N$,

\[\begin{align*} \left\lvert \sum_{k=p}^q a_k b_k \right\rvert &= \left\lvert\sum_{k=p}^{q-1}A_k(b_k - b_{k-1}) - A_{p-1}b_p - A_q b_q \right\rvert \\ &\leq \sum_{k=p}^{q-1}\lvert A_k\rvert \lvert b_k - b_{k+1}\rvert + \lvert A_{p-1}\rvert b_p + \lvert A_q \rvert b_q \\ &\leq M \left( \sum_{k=p}^{q-1}(b_k - b_{k-1}) + b_p +b_q \right) \\ &=2Mb_p \\ &<\epsilon \end{align*}\]

By Theorem 7.2.1 the first equality holds. By Cauchy criterion, $\sum_{k=1}^\infty a_kb_k$ converges.

\[\tag*{$\square$}\]

Theorem 7.2.3 (Alternating Series Test)

If $\{b_k\}$ is a sequence of real numbers satisfying

(a) $b_1 \geq b_2 \geq \cdots \geq 0$, and

(b) $\lim_{k\to\infty} b_k=0$

then $\sum_{k=1}^\infty (-1)^{k+1}b_k$ converges.

<Proof>

Let $a_k =(-1)^{k+1}$ and let $A_n=\sum_{k=1}^na_k$. Then $\lvert A_n \rvert \leq 1$ for all $n\in\mathbb{N}$. By Dirichlet test $\sum_{k=1}^\infty (-1)^{k+1}b_k$ converges.

\[\tag*{$\square$}\]

Theorem (Leibniz’s Theorem on Alternating Series)

$\sum_{n=1}^\infty c_n$ is a series such that

(a) $c_{2m}\leq 0, c_{2m+1} \geq 0$ for all $m\in\mathbb{N}$,

(b) $\lim_{n\to\infty}c_n = 0$,

(c) $\lvert c_1 \rvert \geq \lvert c_2 \rvert \geq \lvert c_3 \rvert \geq \cdots$,

then $\sum_{n=1}^\infty c_n$ converges.

<Proof>

Let $a_n = (-1)^{n+1}$ and let $b_n=\lvert c_n \rvert$. Then $c_n = a_n b_n$. By Dirichlet test, $\sum_{n=1}^\infty c_n$ converges.

\[\tag*{$\square$}\]

Theorem 7.2.4

Consider the series $\sum_{k=1}^\infty (-1)^{k+1}b_k$, where the sequence $\{b_k\}$ satisfies the hypothesis of Theorem 7.2.3. Let

\[\begin{align*} S_n = \sum_{k=1}^n (-1)^{k+1}b_k \quad \text{and} \quad S = \sum_{k=1}^\infty (-1)^{k+1}b_k. \end{align*}\]

Then $\lvert s- s_n\rvert\leq b_{n+1}$ for all $n\in\mathbb{N}$.

<Proof>

Since $\{b_k\}$ is monotone decreasing $b_{k-1}-b_k \geq 0$,

\[\begin{align*} S_{2n} &= \sum_{k=1}^{2n} (-1)^{k+1}b_k \\ &=(b_1 - b_2) + (b_2 - b_3) + \cdots + (b_{2n-1} -b_{2n}). \end{align*}\]

Thus $\{S_{2n}\}$ is a monotone increasing sequence. Similarly, $\{S_{2n-1}\}$ is monotone decreasing sequence. Since $S_n\to S$, we have $S_{2n} \to S$ and $S_{2n-1}\to S$ as $n\to\infty$. In other words,

\[\begin{align*} \sup_{n\in\mathbb{N}} S_{2n} = S = \inf_{n\in\mathbb{N}}S_{2n-1}, \end{align*}\]

which implies that $S_{2n} \leq S \leq S_{2n+1}$ for all $n\in\mathbb{N}$. Thus, $\lvert S- S_{2n}\rvert \leq \lvert S_{2n} -S_{2n-1}\rvert$ and $\lvert S - S_{2n-1}\rvert \leq \lvert S_{2n} - S_{2n-1}\rvert$for all $n\in\mathbb{N}$, i.e.,

\[\begin{align*} \lvert S -S_n \vert \leq |S_{n+1}-S_n\rvert = b_{n+1} \end{align*}\]

for all $n\in\mathbb{N}$.

\[\tag*{$\square$}\]

Definition 7.3.1

A series $\sum_{k=1}^\infty a_k$ of real numbers is said to be absolutely convergence if $\sum_{k=1}^n \lvert a_k\rvert$ converges. The series is said to be conditionally convergent if it is convergent but not absolutely convergent.

Theorem 7.3.3

if $\sum_{k=1}^\infty a_k$ converges absolutely, then $\sum_{k=1}^\infty a_k$ converges and

\[\begin{align*} \left\lvert \sum_{k=1}^\infty a_k \right\rvert \leq \sum_{k=1}^\infty \lvert a_k \rvert. \end{align*}\]

Let $\epsilon >0$ be given. Since $\sum_{k=1}^\infty \lvert a_k\rvert$ converges, by Cauchy criterion there is $N\in\mathbb{N}$ such that

\[\begin{align*} m > n \geq N \Rightarrow \sum_{k=n}^m \lvert a_k \rvert < \epsilon. \end{align*}\]

By triangle inequality,

\[\left\lvert \sum_{k=n}^m a_k \right\rvert \leq \sum_{k=n}^m \lvert a_k \rvert < \epsilon\]

for all $m>n\geq N$. Thus $\sum_{k=1}^\infty a_k$ converges.

Finally,

\[\begin{align*} \left\lvert \sum_{k=1}^\infty a_k\right\rvert &= \left\lvert \lim_{n\to\infty} \sum_{k=1}^n a_k\right\rvert \\ &=\lim_{n\to\infty} \left\lvert \sum_{k=1}^n a_k\right\rvert \quad (\because x\mapsto \lvert x\rvert \text{ is continous}) \\ &=\lim_{n\to\infty}\sum_{k=1}^n \lvert a_k \rvert \\ &=\sum_{k=1}^\infty \lvert a_k \rvert \end{align*}\] \[\tag*{$\square$}\]

Definition 7.3.6

A series $\sum a^\prime_k$ is a rearrangement of the series $\sum a_k$ if there exists a one-to-one function $j$ from $\mathbb{N}$ onto $\mathbb{N}$ such that $a^\prime_k = a_{j(k)}$ for all $k\in\mathbb{N}$.

Theorem 7.3.8

If the series $\sum a_k$ converges absolutely, then every rearrangement of $\sum a_k$ converges to the same sum.

<Proof>

Let $\sum a_k^\prime$ be a rearrangement of $\sum a_k$. Let $\epsilon >0$ be given. Then there exists $N\in\mathbb{N}$ such that

\[\begin{align} m>n\geq N \Rightarrow \sum_{k=n}^m \lvert a_k \rvert < \epsilon. \label{eq:1} \end{align}\]

Suppose that $a^\prime_k = a_{j(k)}$, where $j$ is a one-to-one function of $\mathbb{N}$ onto $\mathbb{N}$. Choose an integer $p (p\geq N)$ such that

\[\begin{align*} \{1,2,\ldots, N\} \subset \{j(1), j(2), \ldots, j(p)\}. \end{align*}\]

Such a $p$ exists since $j$ is bijective. For example, we can set $p=\max\{j^{-1}(1), j^{-1}(2), \ldots, j^{-1}(N) \}$. Let

\[\begin{align*} s_n = \sum_{k=1}^n a_k \quad \text{and}\quad s^\prime_n = \sum_{k=1}^n a^\prime_k. \end{align*}\]

If $n\geq p$,

\[\begin{align*} s_n - s^\prime_n = \sum_{k=1}^n a_k - \sum_{k=1}^n a_{j(k)}. \end{align*}\]

By the choice of $p$, the numbers $a_1, \ldots, a_N$ appear in both sums and consequently cancel. Thus, the only terms remaining will have index $k$ or $j(k)$ greater than $N$. Let $\Lambda$ be the set of indices of the remaining terms. Then

\[\begin{align*} \lvert s_n - s^\prime_n \rvert &\leq \sum_{\lambda \in \Lambda} \lvert a_\lambda \rvert\\ & \leq \sum_{k=\min\Lambda}^{\max\Lambda+1} \lvert a_k \rvert \\ &< \epsilon \quad (\because \max\Lambda \geq \min\Lambda >N \text{ and by equation }\ref{eq:1}) \end{align*}\]

$\therefore \lim_{n\to\infty} s^\prime_n = \lim_{n\to\infty}s_n$.

\[\tag*{$\square$}\]

Reference

• Manfred Stoll, 『Introduction to Real Analysis』