Power Series, Taylor Expansions & Analyticity

sequences and series of complex numbers

A sequence of complex numbers is just an ordered list z_1, z_2, z_3, ... where each z_n lives in the plane rather than on the line. We ask the same question we asked in real calculus: as n grows, do these dots settle down toward some single point z, or do they wander forever? A series is what you get when you decide to add the dots up: the infinite sum z_1 + z_2 + z_3 + ... means the running totals (the partial sums s_N = z_1 + ... + z_N) themselves form a sequence, and the series 'equals' the limit of those running totals if that limit exists.

Convergence in the plane is wonderfully simple because it splits into two ordinary real problems. Write z_n = x_n + i y_n. Then z_n converges to z = x + i y exactly when x_n converges to x AND y_n converges to y — the real and imaginary parts each converge on the number line. Equivalently, z_n converges to z when the distance |z_n - z| shrinks to 0. So everything you know about real limits transfers; the only new flavour is that 'close' now means close as points in the plane, measured by the modulus. A useful sanity check: if the series sum z_n converges, then its terms z_n must shrink to 0 (the n-th term test). That is necessary but, as on the real line, not sufficient.

These ideas are the foundation for everything that follows in this part of the subject. A power series — the local DNA of a holomorphic function — is exactly an infinite series whose terms are complex numbers a_n (z - z_0)^n, one for each fixed z. Asking where such a series converges is asking, for which z does this complex series settle down? So sequences and series are not a dry preliminary; they are the language in which Taylor's theorem and analyticity will be spoken.

Take z_n = (1 + i)^n / n!. The terms shrink fast because n! eventually crushes any |1 + i|^n = (sqrt 2)^n, so |z_n| -> 0 and the sequence converges to 0. In fact the series sum (1 + i)^n / n! converges too — it is the Taylor series for e^z evaluated at z = 1 + i, giving e^(1+i) = e (cos 1 + i sin 1).

Factorial growth in the denominator beats geometric growth of the modulus, so the series converges everywhere — this is why e^z is entire.

A series can have terms going to 0 yet still diverge: sum 1/n diverges even though 1/n -> 0. The n-th term test only ever proves divergence (when terms do not go to 0), never convergence.

Also called
complex series複級數