Chapter 5 Notes

Note exercise 5.19 gives us key results and notation:

\[ \begin{align*} a_j^{+} & \coloneqq \begin{cases} a_j & \text{ if } a_j >0 \\ 0 & \text{ if } a_j \leq 0 \end{cases} , & a_j^{-} & \coloneqq \begin{cases} - a_j & \text{ if } a_j < 0 \\ 0 & \text{ if } a_j \geq 0 \end{cases} \end{align*} \]

Let

\[ \begin{align*} S_n & \coloneqq \sum_{j=1}^{n} a_j, & \lvert S \rvert_n & \coloneqq \sum_{j=1}^{n} \lvert a_j \rvert \\ S_{n}^{+} & \coloneqq \sum_{j=1}^{n} a_{j}^{+}, & S_{n}^{-} & \coloneqq \sum_{j=1}^{n} a_{j}^{-} \\ \end{align*} \]

Therefore

\[ \begin{align*} \lvert a_j \rvert &= a_j^{+} + a_{j}^{-} \\ a_{j} &= a_{j}^{+} + a_{j}^{-} \end{align*} \implies \begin{align*} \lvert S \rvert_{n} &= S_{n}^{+} + S_{n}^{-} \\ S_{n} &= S_{n}^{+} - S_{n}^{-} \end{align*} \]

In 5.19 we showed that if \(\sum_{j=1}^{\infty} a_j\) is convergent but not absolutely convergent then both \(\sum_{j=1}^{\infty} a_{j}^{+}\) and \(\sum_{j=1}^{\infty} a_{j}^{-}\) diverge.

In other words.

\(S_n\) converges and \(\lvert S \rvert_{n}\) not converging implies that both \(S_{n}^{+}\) and \(S_{n}^{-}\) do not converge.

Exercise 5.21

Suppose that \(\sum_{j=1}^{\infty} a_{j}\) is convergent but not absolutely convergent.

Let \(l \in \mathbb{R}\) and,

\[ \begin{align*} S(0) &= \left\{ n \in \mathbb{N} : a_n < 0 \right\}, & T(0) &= \left\{ n \in \mathbb{N} : a_n \geq 0 \right\} \end{align*} \]

We then inductively define \(\sigma : \mathbb{N} \to \mathbb{N}\) for each \(n \in \mathbb{N}\),

let \(\tilde{S}_{n} \coloneqq \sum_{j=1}^{n} a_{\sigma(j)}\) , \(\tilde{S}_{\infty} \coloneqq \lim_{n \to \infty} \tilde{S}_{n}\) and \(\tilde{S}_{0} \coloneqq 0\),

\[ \begin{align*} \text{ if } \tilde{S}_{n-1} < l \\ \sigma \left( n \right) & \coloneqq \min T \left( n-1 \right), & T(n) & \coloneqq T(n-1) \setminus \left\{ \sigma(n) \right\}, & S(n) & \coloneqq S(n-1) \\ \text{ if } \tilde{S}_{n-1} \geq l \\ \sigma \left( n \right) & \coloneqq \min S \left( n-1 \right), & T(n) & \coloneqq T(n-1), & S(n) & \coloneqq S(n-1) \setminus \left\{ \sigma(n) \right\} \end{align*} \]

In other words if the sum is above \(l\) pick the next negative term and vice versa.

[ii] \(\sigma\) is a permutation

Solution

Because neither \(S_{n}^{+}\) or \(S_{n}^{-}\) converge there are infinitely many terms such that \(a_{j}^{+}\) and \(a_{j}^{-}\) are non zero. Therefore both \(S(0), T(0)\) are infinite and the construction is able to pick \(\sigma(n)\) at each stage.

[iii]

Let \(n \in \mathbb{N}\), if,

• \(\kappa > |a_r|\) for all \(r \in S(n) \cup T(n)\)
• \(m > n\) and \(S(m) \neq S(n)\), \(T(m) \neq T(n)\)

show that \(|\sum_{j=1}^{m} a_{\sigma(j)} - l| < \kappa\).

(It may be helpful to consider what happens when \(\sum_{j=1}^{m} a_{\sigma(j)} - l\) changes sign.)

Solution

\(S(m) \neq S(n)\), \(T(m) \neq T(n)\)

\(\iff\)

there exists \(k\) such that \(n<k<m\) and \(a_{\sigma(k)}, a_{\sigma(k+1)}\) have different sign.

Let \(k\) be the greatest interger less than \(m\) where this occurs.

Assume that \(\hat{S}_k \geq l\). The \(<l\) case is a very similar proof.

Therefore by sign change \(\hat{S}_{k-1} < l\)

Notice that

\(S(n) \cup T(n) = \mathbb{N} \setminus \left\{ \sigma(1), \dots \sigma(n) \right\} = \left\{ \sigma(n+1), \sigma(n+2), \dots \right\}\)

There are then two sub cases to consider:

Case 1

\(\hat{S}_{k+1} > l\)

\(\hat{S}_{k-1} < l\)

\(\implies \hat{S}_{k-1} + a_{\sigma(k)} < l + a_{\sigma(k)} \implies \hat{S}_{k} - l < a_{\sigma(k)}\)

So \(0 \leq S_{k} -l < a_{\sigma(k)}\)

Because \(k\) is the last integer \(<m\) where \(a_{\sigma(k)}, a_{\sigma(k+1)}\) have different signs, it holds that \(a_{\sigma(j)} > 0\) for all \(j \in \left\{ k+1, \dots, m \right\}\).

So \(0 \leq \hat{S}_{j} - l \leq \hat{S}_{k} - l < a_{\sigma(k)}\) for all \(j \in \left\{ k+1, \dots, m \right\}\)

Case 2

\(\hat{S}_{k+1} \leq l\)