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Cauchy-Hadamard for complex power series, including zero and infinite radius

Statement

For $\sum_{n\ge0}c_n(z-a)^n$ , set $L:=\limsup_{k\to\infty}|c_{k+1}|^{1/(k+1)}\in[0,+\infty],$ so no $0$ th root occurs, and set $R:=\begin{cases}+\infty,&L=0,\\1/L,&0<L<+\infty,\\0,&L=+\infty.\end{cases}$ Then the series converges absolutely for $|z-a|<R$ and diverges for $|z-a|>R$ ; no assertion is made on $|z-a|=R$ . At $z=a$ it converges to $c_0$ , including when $R=0$ . The conventions and prerequisite facts used below are recorded in Complex series, absolute convergence, complex power series, and radius of convergence, Every absolutely convergent complex series converges, and rearrangements preserve its sum, Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , For finite $L$ : $L = \limsup x_k$ iff for every $\varepsilon > 0$ one has $x_k < L + \varepsilon$ eventually and $x_k > L - \varepsilon$ frequently, Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing.

Facts & Assumptions

Given: The coefficient sequence and a complex $z$ .

[L1]

Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing applies to a real series $\sum_{n\ge1}a_n$ using the defined roots $|a_{k+1}|^{1/(k+1)}$ .

[L2]

For finite $L$ : $L = \limsup x_k$ iff for every $\varepsilon > 0$ one has $x_k < L + \varepsilon$ eventually and $x_k > L - \varepsilon$ frequently states that if a real $L$ is the limit superior of $(b_k)$ , then $b_k>L-\varepsilon$ frequently for every real $\varepsilon>0$ .

[L3]

Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ defines $\limsup b_k$ as the infimum of the extended-real tail suprema $\sup\{b_j:j\ge k\}$ .

[L4]

Every absolutely convergent complex series converges, and rearrangements preserve its sum states that every absolutely convergent complex series converges.

[L5]

Every convergent sequence in a metric space is Cauchy states that every convergent sequence in a metric space is Cauchy.

[L6]

Complex series, absolute convergence, complex power series, and radius of convergence defines the partial sums by $S_0=0$ and $S_{N+1}=S_N+a_N$ , and defines convergence through the complex metric.

Proof

technique · direct

1.1

At $z=a$ , every positive-index term vanishes, so the series converges to $c_0$ .

algebra

1.2

Suppose $z\ne a$ , put $r=|z-a|>0$ , and set $b_k=|c_{k+1}|^{1/(k+1)}$ . For the real modulus tail $d_n=|c_n|r^n$ $(n\ge1)$ , its root family is $d_{k+1}^{1/(k+1)}=b_kr$ .

algebra

2.1

If $r<R$ , then $L$ is finite and $L<1/r$ . Put $\varepsilon=(1/r-L)/2>0$ . The eventual-upper-bound clause of [L2] gives $b_k<L+\varepsilon<1/r$ eventually; by [L3], the limit superior of the root family $b_kr$ is therefore at most $(L+\varepsilon)r<1$ . Hence [L1] gives convergence of the modulus tail. (When $L=+\infty$ , $R=0$ and $r<R$ is impossible.)

L1L2L3step 1.2algebra

2.2

Suppose $0<L<+\infty$ and $r>R=1/L$ . Then $\varepsilon:=L-1/r>0$ , and [L2] gives $b_k>L-\varepsilon=1/r$ frequently. At those arbitrarily large indices, step 1.2 gives $d_{k+1}=(b_kr)^{k+1}>1$ .

L2step 1.2algebra

2.3

Suppose $L=+\infty$ and $r>R=0$ . By [L3], every tail supremum of $(b_k)$ is $+\infty$ ; hence $1/r$ is not an upper bound for any tail, so $b_k>1/r$ frequently. Again $d_{k+1}=(b_kr)^{k+1}>1$ at arbitrarily large indices.

L3step 1.2algebra

3.1

In either divergence case, let $S_N$ be the complex partial sums. If $(S_N)$ converged, [L5] would make it Cauchy; but [L6] gives $d_{\mathbb C}(S_{n+1},S_n)=|c_n(z-a)^n|=d_n>1$ for arbitrarily large $n$ , contradicting the Cauchy condition with tolerance $1$ . Thus the complex series diverges.

L5L6step 2.2step 2.3

3.2

In the case $r<R$ , step 2.1 says that the complex series is absolutely convergent, so it converges by [L4].

L4step 2.1

4.1

Step 1.1 covers the centre, steps 3.1 and 3.2 cover respectively $|z-a|>R$ and $|z-a|<R$ , and none of these arguments asserts anything when $0<L<+\infty$ and $|z-a|=R$ . This proves all three radius cases exactly as stated.

step 1.1step 3.1step 3.2∎

Depends on

Used by

The complex geometric power series has radius 1 and sums to 1/(1-z) for |z|<1 Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 142 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential (standard reference, not scraped)