Math 2.0

Sequences, series, summability Definition

Pringsheim convergence

Builds on Double sequence

Definition

Let (am,n)m,n≥1(a_{m,n})_{m, n \ge 1} be a double sequence of real or complex numbers, and let LL be a number. The double sequence converges to LL in the sense of Pringsheim if the following holds. For every ε>0\varepsilon > 0 there is an integer NN such that ∣am,n−L∣<ε\abs{a_{m,n} - L} < \varepsilon for all m≥Nm \ge N and n≥Nn \ge N. Then LL is unique; it is called the double limit and is written lim⁡m,n→∞am,n=L\lim_{m, n \to \infty} a_{m,n} = L.

Remarks

Example: am,n=1m+1na_{m,n} = \frac{1}{m} + \frac{1}{n} converges to 00, because m,n≥Nm, n \ge N gives ∣am,n∣≤2N\abs{a_{m,n}} \le \frac{2}{N}, which is less than ε\varepsilon once N>2/εN > 2/\varepsilon. Non-example: am,n=mm+na_{m,n} = \frac{m}{m+n} has no double limit, because aN,N=12a_{N,N} = \frac12 while aN,n→0a_{N,n} \to 0 as n→∞n \to \infty, so no single LL works for ε=14\varepsilon = \frac14. A Pringsheim convergent double sequence need not be bounded: put a1,n=na_{1,n} = n and am,n=0a_{m,n} = 0 for m≥2m \ge 2; the double limit is 00, yet the first row diverges. So the double limit may exist while an iterated limit lim⁡mlim⁡nam,n\lim_m \lim_n a_{m,n} does not; the Moore–Osgood theorem gives conditions under which they agree.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Iterated_limit
About this page
Tag4FK55CYMVZ
URIm2://4FK55CYMVZ
Kinddefinition
WritingW0 stub
MSC40B05, 40A05
DefinesPringsheim convergence
UsesDouble sequence
Changed2 November 2026

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