1.12. ROOT TEST 33

Proof: |cn (z)| ≤ ∑nk=0 |an−k (z)| |bk (z)| ≤ ∑

nk=0 An−kBk. Also, from Theorem 1.11.3,

∞

∑n=0

n

∑k=0

An−kBk =∞

∑k=0

∞

∑n=k

An−kBk =∞

∑k=0

Bk

∞

∑n=0

An < ∞.

The claim of 1.12 follows from Merten’s theorem, Theorem 1.11.13. ■

Corollary 1.11.15 Let P be a polynomial and let ∑∞n=0 an (z) converge uniformly and

absolutely on K such that the |an (z)| ≤ An,∑n An < ∞. Then there exists a series forP(∑∞

n=0 an (z)) denoted as ∑∞n=0 cn (z) , which also converges absolutely and uniformly for

z ∈ K because cn (z) also satisfies the conditions of the Weierstrass M test.

1.12 Root TestThe root test has to do with when a series of complex numbers converges. I am assumingthe reader has been exposed to infinite series. However, this that I am about to explain is alittle more general than what is usually seen in calculus.

Theorem 1.12.1 Let ak ∈Fpand consider ∑∞k=1 ak. Then this series converges abso-

lutely if limsupk→∞ |ak|1/k = r < 1. The series diverges spectacularly if limsupk→∞ |ak|1/k >

1 and if limsupk→∞ |ak|1/k = 1, the test fails.

Proof: Suppose first that limsupk→∞ |ak|1/k = r < 1. Then letting R ∈ (r,1) , it followsfrom the definition of limsup that for all k large enough, |ak|1/k ≤ R. Hence there exists Nsuch that if k≥ N, then |ak| ≤ Rk. Let Mk = |ak| for k < N and let Mk = Rk for k≥ N. Then

∞

∑k=1

Mk ≤N−1

∑k=1|ak|+

RN

1−R< ∞

and so, by the Weierstrass M test applied to the series of constants, the series converges andalso converges absolutely. If limsupk→∞ |ak|1/k = r > 1, then letting r > R > 1, it followsthat for infinitely many k, |ak| > Rk and so there is a subsequence which is unbounded.In particular, the series cannot converge and in fact diverges spectacularly. In case thatthe limsup = 1, you can consider ∑

∞n=1

1n which diverges by calculus and ∑

∞n=1

1n2 which

converges, also from calculus. However, the limsup equals 1 for both of these. ■There is no change in the proof if ak is in a complete normed vector space which will

be mentioned later.This is a major theorem because the limsup always exists. As an important application,

here is a corollary which emphasizes one aspect of the above theorem.

Corollary 1.12.2 If ∑k ak converges, then limsupk→∞ |ak|1/k ≤ 1.

If the sequence has values in X a complete normed linear space discussed below, thereis no change in the conclusion or proof of the above theorem. You just replace |·| with ∥·∥the symbol for the norm. Here ∥·∥ is a norm if it satisfies 1.1 - 1.3.