Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product
An absolutely convergent series of complex numbers converges, with the triangle inequality; series dominated by a convergent real series converge absolutely; convergent series add and scale termwise; a series converges exactly when its real and imaginary parts do; a real series converges in the same way in C; and the Cauchy product of two absolutely convergent series converges absolutely to the product of their sums.
Absolute convergence. If ∑k=1∞ak converges absolutely, then it converges, and
k=1∑∞ak≤k=1∑∞∣ak∣.
Comparison. If ∣ak∣≤rk for every k∈N and ∑k=1∞rk converges, then ∑k=1∞ak converges absolutely, and ∑k=1∞ak≤∑k=1∞rk.
Linearity. If ∑k=1∞ak and ∑k=1∞bk converge, then ∑k=1∞(ak+bk) and ∑k=1∞λak converge, with ∑k=1∞(ak+bk)=∑k=1∞ak+∑k=1∞bk and ∑k=1∞λak=λ∑k=1∞ak.
Index shifts. Let p∈N0; ∑k=1∞ap+k is the series of the sequence (ap+k)k∈N (a sequence, as p+k∈N for k∈N), and for p=0 the finite sum ∑k=10ak is the empty sum 0 (Sums and Products over a Finite Set and over an Interval (ii)). Then ∑k=1∞ak converges if and only if ∑k=1∞ap+k converges, and in that case
k=1∑∞ak=k=1∑pak+k=1∑∞ap+k.
Real and imaginary parts.∑k=1∞ak converges if and only if ∑k=1∞Reak and ∑k=1∞Imak converge, and in that case
Rek=1∑∞ak=k=1∑∞Reak,Imk=1∑∞ak=k=1∑∞Imak.
Conjugates. If ∑k=1∞ak converges, then ∑k=1∞ak converges, and
k=1∑∞ak=k=1∑∞ak.
Real series.∑k=1∞ι(rk) converges if and only if ∑k=1∞rk converges, and then ∑k=1∞ι(rk)=ι(∑k=1∞rk).
The Cauchy product. Suppose that ∑k=1∞ak and ∑k=1∞bk converge absolutely, and let (cn) be the sequence in C given by
Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.