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A uniform inverse-sine majorant and geometric angular tail

Statement

Put $h = 3.168$ , $t = 3.326$ , $r = t / h$ and $q = 2.1274$ , with these decimals interpreted as exact rationals. For $0 \leq x \leq 1 / 4$ , $\arcsin x \leq r x$ . Consequently, for every real $C \geq 63.295$ , the series $\sum_{k = 0}^{\infty} \arcsin (h / (C q^{k}))$ converges and its sum is at most $t / (C (1 - 1 / q))$ . For every integer $m \geq 0$ , the tail series also converges and satisfies $0 \leq \sum_{k = m}^{\infty} \arcsin (h / (C q^{k})) \leq t / (C q^{m} (1 - 1 / q))$ .

Facts & Assumptions

[F1]

Rational sine enclosures from a finite factorial sum gives $y - y^{3} / 6 \leq \sin y \leq y$ for rational $y$ with $0 \leq y \leq 1$ ; in particular $\sin 0 = 0$ .

[F2]

Principal inverse sine and inverse cosine supplies sine continuity and strict increase on $[- π / 2, π / 2]$ , and its principal inverse.

[F3]

Rational logarithm and pi enclosures after range reduction supplies a rational enclosure of $π$ ; its $N = 1$ Machin bound gives $π > 3$ .

[F4]

Series, partial sums, convergence and the sum, divergence, and the tail series defines a series sum as the limit of its partial sums and defines the tail starting at an integer $m \geq 0$ as the series with terms $a_{m + j}$ for integers $j \geq 0$ .

[F5]

A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum states that a real series with nonnegative terms converges if its partial sums are bounded above, and its sum is then the supremum of those partial sums.

[F6]

The rationals embed densely in the reals states that strictly between any two distinct real numbers lies a rational number.

Proof

Given: The exact rational constants above, and real $C \geq 63.295$ and integer $m \geq 0$ for the series and tail claims.

Goal: Prove the stated inverse-sine and geometric-tail estimates.

1.1givenF1F2F3F6choosealgebra

We first extend the lower bound in [F1] to all real $y \in [0, 1]$ . Put $p (y) = y - y^{3} / 6$ . For $y, z \in [0, 1]$ , factorization gives $∣ p (z) - p (y) ∣ = ∣ z - y ∣ ∣ 1 - (z^{2} + z y + y^{2}) / 6 ∣ \leq (3 / 2) ∣ z - y ∣$ . The endpoints $y = 0, 1$ are rational, so [F1] applies there. Fix real $0 < y < 1$ and $ε > 0$ . By [F3], $[0, 1] \subset [- π / 2, π / 2]$ , and continuity in [F2] gives $δ > 0$ such that $∣ \sin z - \sin y ∣ < ε / 2$ for branch points $z$ with $∣ z - y ∣ < δ$ . Set $η = \min {δ, ε / 3, y, 1 - y} > 0$ . By [F6], choose a rational $z$ strictly between $y - η$ and $y + η$ . Then $0 < z < 1$ , $∣ p (z) - p (y) ∣ < ε / 2$ , and [F1] gives $p (z) \leq \sin z$ . Therefore $p (y) < p (z) + ε / 2 \leq \sin z + ε / 2 < \sin y + ε$ . Since this holds for every $ε > 0$ , $p (y) \leq \sin y$ : a positive difference would violate it at half that difference. Thus $y - y^{3} / 6 \leq \sin y$ for every real $0 \leq y \leq 1$ .

2.1givenF1F2F3step 1.1algebra

Exact rational arithmetic gives $1 < r < 21 / 20$ and $r - r^{3} / 96 > 1$ . For $0 \leq x \leq 1 / 4$ , $0 \leq r x \leq 21 / 80 < 1 < π / 2$ , and step 1.1 yields $\sin (r x) \geq r x - r^{3} x^{3} / 6 \geq x (r - r^{3} / 96) \geq x$ . By increasing-branch inversion, $\arcsin x \leq r x$ . Since $\sin 0 = 0$ , the same branch inversion gives $\arcsin 0 = 0$ and $\arcsin x \geq 0$ for $x \geq 0$ in this interval. At $x = 0$ equality holds.

3.1givenstep 2.1algebra

Since $h / 63.295 < 1 / 4$ and $q > 1$ , for every integer $k \geq 0$ the argument $h / (C q^{k})$ lies in $[0, 1 / 4]$ . Set $a_{k} = \arcsin (h / (C q^{k}))$ . Step 2.1 gives $0 \leq a_{k} \leq t / (C q^{k})$ . Thus these are well-defined nonnegative real terms.

4.1givenF4step 3.1algebra

Put $ρ = 1 / q$ , so $0 < ρ < 1$ . For integers $m, n \geq 0$ , let $P_{m, n} = \sum_{j = 0}^{n - 1} a_{m + j}$ and $G_{n} = \sum_{j = 0}^{n - 1} ρ^{j}$ , with both sums zero when $n = 0$ . The finite telescoping identity $(1 - ρ) G_{n} = \sum_{j = 0}^{n - 1} (ρ^{j} - ρ^{j + 1}) = 1 - ρ^{n}$ holds also at $n = 0$ . Since $0 < ρ^{n} \leq 1$ , it gives $0 \leq G_{n} \leq 1 / (1 - ρ)$ . Termwise finite summation of step 3.1 therefore gives $0 \leq P_{m, n} \leq (t / (C q^{m})) G_{n} \leq B_{m}$ , where $B_{m} = t / (C q^{m} (1 - 1 / q))$ is positive and independent of $n$ .

5.1givenF4F5step 3.1step 4.1∎

Fix any integer $m \geq 0$ . By [F4], $P_{m, n}$ are precisely the partial sums of the tail series with terms $a_{m + j}$ . These terms are nonnegative by step 3.1, and their partial sums are bounded above by $B_{m}$ by step 4.1. Hence [F5] proves convergence and identifies the tail sum with the supremum of these partial sums, which lies in $[0, B_{m}]$ . This proves the stated tail bound for every integer $m \geq 0$ ; taking $m = 0$ gives convergence and the bound for the full series.

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