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Cauchy–Hadamard: the reciprocal radius is $\limsup_{k\to\infty}|a_{k+1}|^{1/(k+1)}$ , with the zero and infinite cases included

Statement

Let $\sum_{n\ge0}a_n(x-c)^n$ be a real power series with radius $R$ (A real power series about a centre, its interval of convergence, and its radius in $[0,+\infty]$ ), and put

$L:=\limsup_{k\to\infty}|a_{k+1}|^{1/(k+1)}\in[0,+\infty].$

Then $R$ is the reciprocal of $L$ in the following explicit sense:

$R=\begin{cases}+\infty,&L=0,\\[2pt]1/L,&0<L<+\infty,\\[2pt]0,&L=+\infty.\end{cases}$

Equivalently, with the conventions $1/0:=+\infty$ and $1/(+\infty):=0$ , one has $R=1/L$ . The roots use $a_{k+1}$ and the exponent $1/(k+1)$ because $\mathbb N$ starts at $0$ and a zeroth root is undefined.

Facts & Assumptions

Given: A real power series $\sum a_n(x-c)^n$ , its radius $R$ , the nonnegative root sequence $q_k:=|a_{k+1}|^{1/(k+1)}$ , and $L:=\limsup_k q_k$ .

[L1]

The limit superior $L$ exists in $\overline{\mathbb R}$ for every real sequence (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 $\overline{\mathbb{R}}$ , The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence).

[L2]

If $L$ is real, then for every real $\varepsilon>0$ , $q_k<L+\varepsilon$ eventually and $q_k>L-\varepsilon$ frequently (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).

[L3]

The root test says that a real series from index $1$ converges absolutely when the limit superior of its shifted roots is $<1$ , and diverges when that limit superior is $>1$ (Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L4]

Absolute convergence means convergence of the series of absolute values (Absolutely convergent and conditionally convergent series, and the general starting index).

Proof

technique · direct

1.1

Fix $x\in\mathbb R$ and put $d:=|x-c|$ . The shifted roots of the terms $a_n(x-c)^n$ , $n\ge1$ , are $|a_{k+1}(x-c)^{k+1}|^{1/(k+1)}=q_kd$ .

givenalgebra

2.1

If $L=0$ , then for $d=0$ every root in step 1.1 is $0$ , while for $d>0$ and any $\eta>0$ , [L2] applied with $\varepsilon=\eta/d$ makes $q_kd<\eta$ eventually. Thus $\limsup_k(q_kd)=0<1$ for every $x$ .

step 1.1L2

2.2

Suppose $0<L<+\infty$ . If $d<1/L$ , choose a real $t$ with $L<t<1/d$ (with the second inequality omitted when $d=0$ ). By [L2], $q_k<t$ eventually, so $\limsup_k(q_kd)\le td<1$ . If $d>1/L$ , choose $t$ with $1/d<t<L$ ; [L2] gives $q_k>t$ frequently, so $\limsup_k(q_kd)\ge td>1$ .

step 1.1L2choose

2.3

If $L=+\infty$ and $d>0$ , then for every real $M>0$ and every index $N$ there is $k\ge N$ with $q_k>M$ : otherwise $M$ would bound a tail and its supremum, forcing the infimum of the tail suprema to be finite. Taking $M>1/d$ shows $q_kd>1$ arbitrarily late, hence $\limsup_k(q_kd)>1$ .

L1step 1.1choose

3.1

By [L3] and [L4], step 2.1 gives absolute convergence at every real $x$ when $L=0$ ; step 2.2 gives absolute convergence for $d<1/L$ and divergence for $d>1/L$ when $0<L<+\infty$ ; and step 2.3 gives divergence at every $x\ne c$ when $L=+\infty$ , while the series converges at $c$ to $a_0$ .

step 2.1step 2.2step 2.3L3L4

4.1

Reading these three alternatives through the definition of the radius yields $R=+\infty$ , $R=1/L$ , and $R=0$ , respectively, which is the stated convention-complete formula.

step 3.1∎

Depends on

Used by

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Sources

Cauchy-Hadamard theorem, Encyclopedia of Mathematics (standard reference, not scraped)
Power series, Encyclopedia of Mathematics (standard reference, not scraped)

Cauchy–Hadamard: the reciprocal radius is lim sup⁡k→∞∣ak+1∣1/(k+1)\limsup_{k\to\infty}|a_{k+1}|^{1/(k+1)}limsupk→∞​∣ak+1​∣1/(k+1), with the zero and infinite cases included