Math 2.0

Sequences, series, summability Definition

Matrix summability method

Builds on Summability methodSequenceSeries

Definition

Let A=(ank)n,k≥0A = (a_{nk})_{n, k \ge 0} be an infinite matrix of complex numbers. For a complex sequence s=(sk)k≥0s = (s_k)_{k \ge 0}, the AA-transform of ss is the sequence (As)n=∑k=0∞anksk(n≥0),(As)_n = \sum_{k=0}^{\infty} a_{nk} s_k \qquad (n \ge 0), defined when each of these series converges. The sequence ss is AA-summable to LL if (As)n(As)_n is defined for every nn and (As)n→L(As)_n \to L as n→∞n \to \infty. The resulting summability method, with DAD_A the set of AA-summable sequences and A(s)=lim⁡n(As)nA(s) = \lim_n (As)_n, is the matrix method given by AA.

Remarks

Example: the Cesàro matrix has ank=1n+1a_{nk} = \frac{1}{n+1} for k≤nk \le n and ank=0a_{nk} = 0 for k>nk > n. For s=(1,0,1,0,… )s = (1, 0, 1, 0, \dots) one has (As)n=⌈(n+1)/2⌉n+1→12(As)_n = \frac{\lceil (n+1)/2 \rceil}{n+1} \to \frac12, so ss is AA-summable to 12\frac12. Non-example: for ank=2−k−1a_{nk} = 2^{-k-1} and sk=2ks_k = 2^k, the series ∑kanksk=∑k12\sum_k a_{nk} s_k = \sum_k \frac12 diverges, so (As)n(As)_n is not defined and ss is not AA-summable. The identity matrix gives ordinary convergence. Every matrix method is linear, because A(λs+μt)=λAs+μAtA(\lambda s + \mu t) = \lambda As + \mu At.

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  1. Sequence definition
  2. Real sequence definition
  3. Series definition
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  6. Matrix summability method definition

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Further reading

  1. Web https://en.wikipedia.org/wiki/Silverman–Toeplitz_theorem
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MSC40C05
DefinesMatrix summability method
UsesSummability method, Sequence, Series
Changed2 November 2026

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