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} & b_{0,1} & b_{0,2} & \cdots \\ 0 & b_{1,0} & b_{1,1} & \cdots \\ 0 & 0 & b_{2,0} & \cdots \\ \vdots & \vdots & \vdots & \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,&a_n\ge0\\0,&a_n<0\end{cases} and q_n=a_n^-=\begin{cases}0,&a_n\ge0\\-a_n,&a_n<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."