Double Series

When can we assume row sum, col sum and diagsum are all the same in a double series?

When all terms are non negative ($\ge0$ )

What does it mean for a series Cn to be a rearrangement of An

there exists a bijective function $f:\mathbb{N}\cup\left\lbrace{0}\right\rbrace\to\mathbb{N}\cup\left\lbrace{0}\right\rbrace$

such that $c_{n}=a_{f(n)}$

Whats the nullity test for divergence?

If $a_{n}\not\rightarrow 0$ as $n→\infty$, the series is divergent.

Whats the algebra of infinite sums?

If $a_n$ is convergent with sum s and $b_n$ is convergent with sum T, then $\forall\lambda,\mu\in\mathbb{R}$ $\sum_{n=1}^{\infty}\left(\lambda a_{n}+\mu b_{n}\right)$ is convergent with sum $\lambda S+\mu T$

What does it mean for a series to be absolutely convergent?

If \sum_{n=1}^{\infty}\left|an\right|<+\infty

What is the absolute convergence theorem and the Infinite Triangle Inequality?

Suppose that $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent. Then:

1. there are non-negative $p_n,q_n$ such that \sum_{n=1}^{\infty}p_{n}<+\infty,\sum_{n=1}^{\infty}q_{n}<+\infty and $a_n=p_n-q_n$ for all n.

2. $\sum_{n-1}^{\infty}a_{n}$ is convergent

3. $\left|\sum_{n=1}^{\infty}a_{n}\right|\le\sum_{n=1}^{\infty}\left|a_{n}\right|$ (infinite triangle inequality).

Whats a conditionally convergent series?

Convergent but not absolutely convergent

Whats the alternating series test?

Let $a_1,a_2,…\ge0$ be a decreasing sequence which tends to 0. Then the series $a_1-a_2+a_3-a_4+\ldots=\sum_{n=1}^{\infty}\left(-1\right)^{n+1}a_{n}$ is convergent.

Proof that ColSum, RowSum and DiagSum are all equal for a non-negative double series

Let $s_{p,q}$ be the sum of the finite $(p+1)\times(q+1)$ rectangle. Looking at columns, each column sum inside the rectangle is at most the full column sum, so $s_{p,q} \le \text{ColSum}_0+\cdots+\text{ColSum}_q \le C$.

Now fix $p$ and let $q\to\infty$. Then $s_{p,q}\to \sum_{m=0}^p \text{RowSum}_m$, so $\sum_{m=0}^p \text{RowSum}_m \le C$ for every $p$.

Thus all partial sums of $R$ are bounded above by $C$.

Since the terms are non-negative, the Upper Bound Test gives $R \le C$.

Next apply exactly the same argument to the transposed array $b'_{m,n}=b_{n,m}$. This swaps rows and columns, so we get $C \le R$. Hence $R=C$.

Finally, form the shifted array $\begin{pmatrix} b_{0,0} &amp; b_{0,1} &amp; b_{0,2} &amp; \cdots \\ 0 &amp; b_{1,0} &amp; b_{1,1} &amp; \cdots \\ 0 &amp; 0 &amp; b_{2,0} &amp; \cdots \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{pmatrix}$. Its row sums are the same as the original row sums, but its column sums are exactly the diagonal sums of the original array. Applying the already proved result $R=C$ to this shifted array gives $R=D$. Therefore $R=C=D$."

State and prove the Rearrangement Theorem for non-negative series.

Statement: If $a_n \ge 0$ for all $n$, then any rearrangement $\sum_{n=0}^\infty a_{\sigma(n)}$ has the same sum as $\sum_{n=0}^\infty a_n$.

Proof: Let $\sigma:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a bijection and define a double series by $b_{\sigma(n),n}=a_{\sigma(n)}$ and $b_{m,n}=0$ if $m\ne \sigma(n)$.

In each row $m$, exactly one term is non-zero (since $\sigma$ is bijective), and it equals $a_m$, so $\text{RowSum}_m=a_m$ and hence $R=\sum_{m=0}^\infty a_m$ (original sum).

In each column $n$, the only non-zero term is $b_{\sigma(n),n}=a_{\sigma(n)}$, so $\text{ColSum}_n=a_{\sigma(n)}$ and hence $C=\sum_{n=0}^\infty a_{\sigma(n)}$ (rearranged sum). Since all $b_{m,n}\ge 0$ , we can change the order of summation and get $R=C$. Therefore the rearranged series has the same sum as the original.

Proof of the nullity test for divergence?

Proof: Let $s_n = a_1 + \cdots + a_n$ be the partial sums.

Suppose for contradiction that $\sum_{n=1}^\infty a_n$ is convergent, so $s_n \to s$ for some real number $s$. Then also $s_{n-1} \to s$. By algebra of limits, $s_n - s_{n-1} \to s - s = 0$. But $s_n - s_{n-1} = a_n$, so $a_n \to 0$.

This contradicts the assumption that $a_n \not\to 0$. Therefore the series must be divergent.

Proof of the absolute convergence theorem

Statement: If \sum_{n=1}^\infty |a_n| < \infty, then:

$1.$ there exist non-negative sequences $p_n,q_n$ such that $a_n=p_n-q_n$ for all $n$ and both $\sum_{n=1}^\infty p_n$ and $\sum_{n=1}^\infty q_n$ converge;

$2.$ the series $\sum_{n=1}^\infty a_n$ converges;

$3.$ $\left|\sum_{n=1}^\infty a_n\right| \le \sum_{n=1}^\infty |a_n|$.

Proof: Define $p_n=a_n^+=\begin{cases}a_n,&amp;a_n\ge0\\0,&amp;a_n&lt;0\end{cases}$ and $q_n=a_n^-=\begin{cases}0,&amp;a_n\ge0\\-a_n,&amp;a_n&lt;0\end{cases}$.

Then $p_n,q_n\ge0$ and $a_n=p_n-q_n,\quad |a_n|=p_n+q_n,\quad 0\le p_n,q_n\le |a_n|$.

Since $\sum |a_n|$ converges, the Comparison Test gives that $\sum p_n$ and $\sum q_n$ both converge, say to $P$ and $Q$.

By Algebra of Infinite Sums, $\sum a_n=\sum (p_n-q_n)$ converges and has sum $P-Q$. Also, because $0\le P,Q\le \sum |a_n|=:M$, we have $|P-Q|\le M$.

Therefore $\left|\sum_{n=1}^\infty a_n\right|=|P-Q|\le \sum_{n=1}^\infty |a_n|$."

Proof of the alternating series test.

Statement: Let $a_1,a_2,\dots \ge 0$ be a decreasing sequence with $a_n \to 0$. Then the alternating series $a_1-a_2+a_3-a_4+\cdots=\sum_{n=1}^\infty (-1)^{n+1}a_n$ is convergent.

Proof: Write the series as the sum of two series: Series 1: $(a_1-a_2)+0+(a_3-a_4)+0+(a_5-a_6)+0+\cdots$ and Series 2: $a_2-a_2+a_4-a_4+a_6-a_6+\cdots$.

For Series 1, the terms are non-negative because $a_n$ is decreasing, so $a_1-a_2\ge0,\ a_3-a_4\ge0,\dots$.

Its partial sums are bounded above by $a_1$, since each step subtracts at least as much as the next addition gives back.

So by the Upper Bound Test, Series 1 converges.

For Series 2, the partial sums are $a_2,0,a_4,0,a_6,0,\dots$.

Each partial sum lies between $0$ and $a_n$, and since $a_n\to0$, the Sandwich Rule gives that these partial sums tend to $0$.

So Series 2 converges.

The original alternating series is Series 1 plus Series 2, so by Algebra of Infinite Sums it is convergent."