Math 2.0

Sequences, series, summability Definition

Double series

Builds on Double sequencePringsheim convergenceComplex numbers

Definition

Let (am,n)m,n∈ℕ(a_{m,n})_{m, n \in \N} be a double sequence of complex numbers. The double series ∑m,n=0∞am,n\sum_{m, n = 0}^{\infty} a_{m,n} is the double sequence of rectangular partial sums SM,N=∑m=0M∑n=0Nam,n.S_{M,N} = \sum_{m=0}^{M} \sum_{n=0}^{N} a_{m,n}. The double series converges to s∈ℂs \in \C if (SM,N)(S_{M,N}) converges to ss in the sense of Pringsheim. That is, for every ε>0\varepsilon > 0 there is KK with ∣SM,N−s∣<ε\abs{S_{M,N} - s} < \varepsilon for all M,N≥KM, N \ge K.

Remarks

Example: for am,n=2−m3−na_{m,n} = 2^{-m} 3^{-n} one has SM,N=(∑m≤M2−m)(∑n≤N3−n)S_{M,N} = \left(\sum_{m \le M} 2^{-m}\right)\left(\sum_{n \le N} 3^{-n}\right). The two factors tend to 22 and 32\tfrac32, so the double series converges to 33. Non-example: for am,n=1a_{m,n} = 1 when m=nm = n and 00 otherwise, SM,N=min⁡(M,N)+1S_{M,N} = \min(M, N) + 1, which is unbounded, so the double series diverges. A convergent double series can have unbounded terms. With a0,n=na_{0,n} = n, a1,n=−na_{1,n} = -n and am,n=0a_{m,n} = 0 for m≥2m \ge 2, one has SM,N=0S_{M,N} = 0 for all M≥1M \ge 1. So it converges to 00, although its rows do not form convergent series.

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  1. Complex numbers object
  2. Function concept
  3. Pringsheim convergence concept
  4. Sequence definition
  5. Double sequence definition
  6. Double series definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Series_(mathematics)
About this page
TagCWP26Y1TCZ
URIm2://CWP26Y1TCZ
Kinddefinition
WritingW0 stub
MSC40B05
DefinesDouble series
UsesDouble sequence, Pringsheim convergence, Complex numbers
Changed2 November 2026

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