linguistic:voice_change
Proof that any absolutely convergent series of complex numbers is convergent

Suppose that ∑ | a k |, a k ∈ C {\textstyle \sum \left|a_{k}\right|,a_{k}\in \mathbb {C} } is convergent. Then equivalently, ∑ 1 / 2 {\textstyle \sum \left^{1/2}} is convergent, which implies that ∑ | Re (a k) | {\textstyle \sum \left|\operatorname {Re} \left(a_{k}\right)\right|} and ∑ | Im (a k) | {\textstyle \sum \left|\operatorname {Im} \left(a_{k}\right)\right|} converge by termwise comparison of non-negative terms. It suffices to show that the convergence of these series implies the convergence of ∑ Re (a k) {\textstyle \sum \operatorname {Re} \left(a_{k}\right)} and ∑ Im (a k), {\textstyle \sum \operatorname {Im} \left(a_{k}\right),} for then, the convergence of ∑ a k = ∑ Re (a k) + i ∑ Im (a k) {\textstyle \sum a_{k}=\sum \operatorname {Re} \left(a_{k}\right)+i\sum \operatorname {Im} \left(a_{k}\right)} would follow, by the definition of the convergence of complex-valued series.

The preceding discussion shows that we need only prove that convergence of ∑ | a k |, a k ∈ R {\textstyle \sum \left|a_{k}\right|,a_{k}\in \mathbb {R} } implies the convergence of ∑ a k. {\textstyle \sum a_{k}.}

Let ∑ | a k |, a k ∈ R {\textstyle \sum \left|a_{k}\right|,a_{k}\in \mathbb {R} } be convergent. Since 0 ≤ a k + | a k | ≤ 2 | a k |, {\displaystyle 0\leq a_{k}+\left|a_{k}\right|\leq 2\left|a_{k}\right|,} we have 0 ≤ ∑ k = 1 n (a k + | a k |) ≤ ∑ k = 1 n 2 | a k |. {\displaystyle 0\leq \sum _{k=1}^{n}(a_{k}+\left|a_{k}\right|)\leq \sum _{k=1}^{n}2\left|a_{k}\right|.} Since ∑ 2 | a k | {\textstyle \sum 2\left|a_{k}\right|} is convergent, s n = ∑ k = 1 n (a k + | a k |) {\textstyle s_{n}=\sum _{k=1}^{n}\left(a_{k}+\left|a_{k}\right|\right)} is a bounded monotonic sequence of partial sums, and ∑ (a k + | a k |) {\textstyle \sum \left(a_{k}+\left|a_{k}\right|\right)} must also converge. Noting that ∑ a k = ∑ (a k + | a k |) − ∑ | a k | {\textstyle \sum a_{k}=\sum \left(a_{k}+\left|a_{k}\right|\right)-\sum \left|a_{k}\right|} is the difference of convergent series, we conclude that it too is a convergent series, as desired.