Proof. We have

\[ s_{2n}-s_n = \frac 1{n+1}+\frac 1{n+2}+\dots +\frac 1{2n} \ge n\times \frac 1{2n}=\frac 12. \]

Hence \(s_2\ge s_1+\frac 12,\) \(s_4\ge s_2+\frac 12\ge s_1+\frac 22,\) and, continuing, we get \(s_{2^n}\ge s_1+\frac n2.\) This shows that the sequence \((s_n)_{n\ge 1}\) is unbounded. □

Proof. Let \(\sigma \colon \NN \to \NN \) be a bijection so that \(a_{\sigma (1)}+a_{\sigma (2)}+\dots \) is a rearrangement of the original series. We let

\[ s_n = a_1+a_2 + \dots + a_n, \quad t_n = a_{\sigma (1)}+a_{\sigma (2)}+\dots + a_{\sigma (n)} \]

be the \(n\)th partial sum of the series, respectively, rearranged series. Denote by \(M(n)\) the largest among the indices \(\sigma (1), \dots , \sigma (n).\) Then \(a_{\sigma (1)}, a_{\sigma (2)}, \dots , a_{\sigma (n)}\) is a sublist of the list \(a_1,\dots ,a_{M(n)}\) of non-negative real numbers, and so

\[ t_n \le s_{M(n)}. \]

If \(\sum _{n=1}^\infty a_n\) is convergent with sum \(S < +\infty ,\) then \(s_{M(n)} \le S,\) and so \(t_n\le S,\) for all \(n.\) By the boundedness test, Theorem 1.4, the rearranged series is convergent with

\[ \sum \limits _{n=1}^\infty a_{\sigma (n)} \le \sum \limits _{n=1}^\infty a_n, \]

which proves that rearranging a non-negative series cannot increase its sum.

But the series \(\sum _{n=1}^\infty a_n\) can also be viewed as a rearrangement of \(\sum _{n=1}^\infty a_{\sigma (n)},\) via the bijective function \(\sigma ^{-1}.\) Since rearranging cannot increase the sum, we must conclude that

\[ \sum \limits _{n=1}^\infty a_n \le \sum \limits _{n=1}^\infty a_{\sigma (n)} . \]

From the two inequalities we conclude that the series and its rearrangement have the same sum. Finally, our observation that \(\sum _{n=1}^\infty a_n\) is a rearrangement of \(\sum _{n=1}^\infty a_{\sigma (n)}\) proves, by contrapositive, the implication “\(\sum _{n=1}^\infty a_n\) is divergent \(\Rightarrow \) \(\sum _{n=1}^\infty a_{\sigma (n)}\) is divergent”. □

Proof. 1. All the different ways of arranging the terms \(a_{m,n}\) in a single series are rearrangements of one another, hence must have the same sum, \(S,\) by Theorem 2.2.

2. To show that \(\sum \limits _{d=0}^\infty \mathrm {DiagSum}_d = S\) and \(\lim \limits _{b\to \infty } \mathrm {SquareSum}_{b\times b} = S,\) we consider two special ways to enumerate the \(a_{m,n},\) shown in Figure 2.2.

In Figure 2.2(A), partial sums of the resulting single series (which, by part 1., converge to \(S\)) contain a subsequence of sums of the form \(\mathrm {DiagSum}_0+\mathrm {DiagSum}_1+\dots +\mathrm {DiagSum}_d.\) All subsequences of a convergent sequence have the same limit, so this subsequence must also converge to \(S.\)

Figure 2.2(B) similarly shows that \(\lim _{b\to \infty } \mathrm {SquareSum}_{b\times b}\) must be \(S.\)

3. The top row of the \(b\times b\) square sum is \(a_{00}+a_{01}+\dots +a_{0b}.\) This is a partial sum which is less than or equal to the infinite sum \(\mathrm {RowSum}_0.\) In the same way, the remaining rows of \(\mathrm {SquareSum}_{b\times b}\) are bounded above by \(\mathrm {RowSum}_1,\dots ,\mathrm {RowSum}_b.\) Therefore,

\[ \mathrm {SquareSum}_{b\times b} \le \sum _{m=0}^b \mathrm {RowSum}_m. \]

Taking the limit as \(b\to \infty ,\) we have

\[ S = \lim _{b\to \infty } \mathrm {SquareSum}_{b\times b} \le \sum _{m=0}^\infty \mathrm {RowSum}_m. \]

It remains to show that the opposite inequality, \(\sum _{m=0}^\infty \mathrm {RowSum}_m \le S,\) also holds. Since the infinite sum \(\sum _{m=0}^\infty \mathrm {RowSum}_m\) is the limit of partial sums, it is enough to show that for every fixed \(M,\) the sum \(\sum _{m=0}^M \mathrm {RowSum}_m\) is at most \(S.\)

By definition of the row sums, we have

Adding together these limits, we have

\[ \sum _{m=0}^M \mathrm {RowSum}_m = \lim _{n\to \infty } \bigl ( \text {\textrm {sum over the $M\times n$ rectangle}} \bigr ). \]

Yet any rectangle is contained in a \(b\times b\) square for large enough \(b,\) and by part 2., sums over all squares are \(\le S.\) Hence \(\sum _{m=0}^M \mathrm {RowSum}_m\) is a limit of a sequence bounded above by \(S,\) which means that \(\sum _{m=0}^M \mathrm {RowSum}_m \le S.\)

We have proved that \(S \le \sum _{m=0}^\infty \mathrm {RowSum}_m \le S,\) so \(\sum _{m=0}^\infty \mathrm {RowSum}_m =S.\) Finally, the argument for the column sums, \(\sum _{n=0}^\infty \mathrm {ColumnSum}_n = S,\) is the same as for row sums, and we omit it. The Proposition is proved. □

Proof. Write \(s_n\) for the \(n\)th partial sum. Assume that the series is convergent, i.e., \(\lim \limits _{n\to \infty } s_n=s\) for some real number \(s.\) Then \(\lim \limits _{n\to \infty } s_{n-1}=s\) as well. By AoL of convergent sequences, \(\lim _{n\to \infty } (s_n-s_{n-1}) = s-s=0.\) But \(s_n-s_{n-1}=a_n\) and so \(\lim _{n\to \infty } a_n=0.\) We proved:

the series \(\sum \limits _{n=1}^\infty a_n\) is convergent  \(\Rightarrow \)  \(\lim \limits _{n\to \infty }a_n=0,\)

which is the contrapositive of, hence is equivalent to, the statement of the Theorem. □

Proof. The \(n\)th partial sum of the series \(\sum _{n=1}^\infty (\lambda a_n+\mu b_n)\) is \((\lambda a_1+\mu b_1)+\dots + (\lambda a_n+\mu b_n).\) This is a finite sum, so we can rearrange to get \(\lambda s_n+\mu t_n,\) where \(s_n=a_1+\dots +a_n\) and \(t_n=b_1+\dots +b_n\) (partial sums). We are given that \(s_n\to S\) and \(t_n\to T\) as \(n\to \infty ,\) so by AoL of convergent sequences, \(\lambda s_n + \mu t_n\to \lambda S + \mu T\) as claimed. □

Proof. For a real number \(a,\) denote

\(a^+=\begin {cases}a,&\textsf { if }a\ge 0,\\ 0,&\textsf { if }a<0,\end {cases}\)  \(a^- = \begin {cases} 0, &\textsf { if } a\ge 0, \\ -a, &\textsf { if }a<0.\end {cases}\)

1. Assume that the series \(\sum a_n\) is absolutely convergent, meaning that \(\sum _{n=1}^\infty |a_n|\) has finite sum \(M <+\infty .\) Put \(p_n=a_n^+\) and \(q_n = a_n^-\) for all \(n.\) Then we have

\[ 0\le p_n,q_n\le |a_n|, \quad a_n=p_n - q_n, \quad |a_n|=p_n+q_n. \]

By Comparison Test, Theorem 1.5, the series \(\sum _{n=1}^\infty p_n\) is convergent, with some finite sum \(P\le M,\) and likewise \(\sum _{n=1}^\infty q_n= Q \le M.\)

2. By AoIS, the series \(\sum _{n=1}^\infty a_n=\sum _{n=1}^\infty (p_n -q_n)\) is convergent with sum \(P-Q.\)

3. Since \(0\le P,Q\le M,\) we have \(|P-Q|\le M,\) which reads \(\left |\sum _{n=1}^\infty a_n\right |\le \sum _{n=1}^\infty |a_n|.\) □

Proof. Similarly to the Absolute Convergence Theorem, the idea is to write the given series as a linear combination of two convergent series; yet in this case, we cannot use two series with non-negative terms. Consider two series:

\[ \begin {array}{lccccccc} \text {Series 1:} & (a_1-a_2) & +0 & +(a_3-a_4) & +0 & +(a_5-a_6) & +0 & +\dots \\ \text {Series 2:} & a_2 & -a_2 & +a_4 & -a_4& +a_6 & -a_6& +\dots \end {array} \]

To obtain a partial sum of Series 1, we start with \(a_1,\) subtract \(a_2,\) add \(a_3,\) subtract \(a_4,\) add \(a_5\) etc. By assumption, \(a_2, a_3, \dots \) decrease, so each time we subtract more than we add; hence all partial sums are bounded above by \(a_1.\) Series 1 has non-negative terms, hence is convergent by Boundedness Test, Theorem 1.4.

Series 2 has partial sums \(a_2,\) \(0,\) \(a_4,\) \(0,\) \(a_6,\) \(0\) and so on. The \(n\)th partial sum is between \(0\) and \(a_n,\) and \(a_n\to 0\) as \(n\to \infty .\) By Sandwich Rule the partial sums have limit \(0,\) hence Series 2 is convergent.

The required series \(a_1-a_2+a_3-a_4+\dots \) is obtained by adding Series 2 to Series 1, hence is convergent by Algebra of Infinite Sums, Proposition 2.5. □