Questions tagged [double-sequence]

A double sequence is a map from $\mathbb{N}\times \mathbb{N}$ into a space; for instance, a double real sequence $(a_{ij})$ is a map $a:\mathbb{N}\times\mathbb{N}\to \mathbb{R}$. As with the single-variable case, the notation ${a_{ij}}$ is also common.

Double sequences are extensions of sequences that contain two index sets. It is not uncommon to list the first few terms in an array or a matrix; however, these sequences do not possess a matrix structure. Common questions include the evaluation of limits, i.e. conditions when $$ \lim_{m\to\infty}\left(\lim_{n\to \infty} a_{m,n}\right)=\lim_{n\to\infty}\left(\lim_{m\to \infty} a_{m,n}\right)=\lim_{m,n\to\infty}\left( a_{m,n}\right) $$and summability (when the corresponding series converge). A classic technique in the theory is interchanging the order of summation or converting single sums to double sums. For instance, if $\zeta(s)$ is the Riemann zeta function, to evaluate $\sum_{n=2}^{\infty} \zeta(n)-1$ we could write $$ \sum_{n=2}^{\infty} \zeta(n)-1 = \sum_{n=2}^{\infty}\left(\sum_{m=2}^{\infty}\frac{1}{m^n}\right) $$ $$ =\sum_{m=2}^{\infty}\left(\sum_{n=2}^{\infty}\frac{1}{m^n}\right)=\sum_{m=2}^{\infty}\frac{1}{m(m-1)}=1 $$Here the interchange is justified by the non-negativity of the terms.

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Prove absolutely convergent double series $\sum u_{\mu,\nu}$ when rearranged has the same limit.

Prove absolutely convergent double series $\sum u_{\mu,\nu}$ (which tends to S) when rearranged has the same limit. (1) First prove it is true for absolutely convergent series whose items are all non-negative. Say the rearranged series is $\sum…
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Double series summation $S_{\mu, \nu}=\frac{\mu-\nu}{\mu+\nu}$

The double series $\sum u_{\mu, \nu}=\sum\frac{\mu-\nu}{\mu+\nu}$ is said to have sum by rows equaling -1, sum by columns equaling 1. But if we sum by rows, for each row, we have $\sum_{\nu=0}^\infty u_{\mu,…