section_path: Generalizations/Summations over arbitrary index sets/Unconditionally convergent series
section_name: Unconditionally convergent series
paragraph_text: Unconditionally convergent series
  
  Suppose that I = N. {\displaystyle I=\mathbb {N}.} If a family a n, n ∈ N, {\displaystyle a_{n},n\in \mathbb {N},} is unconditionally summable in a Hausdorff abelian topological group X, {\displaystyle X,} then the series in the usual sense converges and has the same sum,
  
  ∑ n = 0 ∞ a n = ∑ n ∈ N a n. {\displaystyle \sum _{n=0}^{\infty }a_{n}=\sum _{n\in \mathbb {N} }a_{n}.}
  
  By nature, the definition of unconditional summability is insensitive to the order of the summation. When ∑ a n {\displaystyle \sum a_{n}} is unconditionally summable, then the series remains convergent after any permutation σ: N → N {\displaystyle \sigma :\mathbb {N} \to \mathbb {N} } of the set N {\displaystyle \mathbb {N} } of indices, with the same sum,
  
  ∑ n = 0 ∞ a σ (n) = ∑ n = 0 ∞ a n. {\displaystyle \sum _{n=0}^{\infty }a_{\sigma (n)}=\sum _{n=0}^{\infty }a_{n}.}
  
  Conversely, if every permutation of a series ∑ a n {\displaystyle \sum a_{n}} converges, then the series is unconditionally convergent. When X {\displaystyle X} is complete then unconditional convergence is also equivalent to the fact that all subseries are convergent; if X {\displaystyle X} is a Banach space, this is equivalent to say that for every sequence of signs ε n = ± 1 {\displaystyle \varepsilon _{n}=\pm 1}, the series
  
  ∑ n = 0 ∞ ε n a n {\displaystyle \sum _{n=0}^{\infty }\varepsilon _{n}a_{n}}
  
  converges in X. {\displaystyle X.}