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

Statement

Let $\sum_{n \geq 0} a_{n} (x - c)^{n}$ 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

$L : = {lim sup}_{k \to \infty} ∣ a_{k + 1} ∣^{1 / (k + 1)} \in [0, + \infty] .$

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

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

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 $N$ 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 $L : = {lim sup}_{k} q_{k}$ .

[L1]

The limit superior $L$ exists in $\overline{R}$ for every real sequence (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L2]

If $L$ is real, then for every real $ε > 0$ , $q_{k} < L + ε$ eventually and $q_{k} > L - ε$ frequently (For finite $L$ : $L = lim sup x_{k}$ iff for every $ε > 0$ one has $x_{k} < L + ε$ eventually and $x_{k} > L - ε$ 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: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ 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

1.1givenalgebra

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

2.1step 1.1L2

If $L = 0$ , then for $d = 0$ every root in step 1.1 is $0$ , while for $d > 0$ and any $η > 0$ , [L2] applied with $ε = η / d$ makes $q_{k} d < η$ eventually. Thus ${lim sup}_{k} (q_{k} d) = 0 < 1$ for every $x$ .

2.2step 1.1L2choose

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 ${lim sup}_{k} (q_{k} d) \leq t d < 1$ . If $d > 1 / L$ , choose $t$ with $1 / d < t < L$ ; [L2] gives $q_{k} > t$ frequently, so ${lim sup}_{k} (q_{k} d) \geq t d > 1$ .

2.3L1step 1.1choose

If $L = + \infty$ and $d > 0$ , then for every real $M > 0$ and every index $N$ there is $k \geq N$ 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 $q_{k} d > 1$ arbitrarily late, hence ${lim sup}_{k} (q_{k} d) > 1$ .

3.1step 2.1step 2.2step 2.3L3L4

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 $x \neq c$ when $L = + \infty$ , while the series converges at $c$ to $a_{0}$ .

4.1step 3.1∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Cauchy–Hadamard: the reciprocal radius is lim sup⁡k→∞∣ak+1∣1/(k+1)limsupk→∞​∣ak+1​∣1/(k+1), with the zero and infinite cases included

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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