Proof of Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named. Natural numbers are read in $\mathbb{R}$ through the canonical map, as fixed in The Real Numbers: Standing Notation and Background (i); by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field every natural number is positive as a real number, and by claim 1 of that lemma the real number attached to $j+1$ is the real number attached to $j$ increased by $1$. We use throughout that $\le$ on $\mathbb{R}$ is a total order, so reflexive (axiom 1) and transitive (axiom 3), and that $x\le y$ holds if and only if $0\le y-x$, by claim 3 of Elementary Arithmetic in an Ordered Field; adding two inequalities $x\le y$ and $u\le v$ to get $x+u\le y+v$ uses that claim together with claim 2 of the same lemma, applied to $(y-x)+(v-u)=(y+v)-(x+u)$.

Claim 1 (Clause 1). Let $0<a$ and $a\le b$. By claim 7 of Elementary Order Arithmetic in an Ordered Field the inverse $a^{-1}$ exists and $0<a^{-1}$. By the mixed transitivity of claim 2 of that lemma, $0<a$ and $a\le b$ give $0<b$, so by claim 7 again $b^{-1}$ exists and $0<b^{-1}$. The product $a^{-1}b^{-1}$ is positive by claim 5 of that lemma, so in particular $0\le a^{-1}b^{-1}$. Applying claim 5 of Elementary Arithmetic in an Ordered Field to $a\le b$ with the nonnegative multiplier $a^{-1}b^{-1}$ gives

$$a^{-1}b^{-1}a\le a^{-1}b^{-1}b .$$

By the commutativity and associativity of multiplication and $a^{-1}a=1$, $b^{-1}b=1$, the left side is $b^{-1}$ and the right side is $a^{-1}$. Hence $b^{-1}\le a^{-1}$.

Claim 2 (Clause 2). Let $P$ be the set of those $M\in\mathbb{N}$ for which $\sum_{m=1}^{M}\tfrac{1}{m^{2}}\le2-\tfrac{1}{M}$; we show $P=\mathbb{N}$ by Principle of Induction for the Natural Numbers with the inductive set $P$.

First, $1\in P$: by claim 1 of Properties of Finite Sums the sum over $[1]$ is its single term, which is $\tfrac{1}{1^{2}}=1$ because $1^{2}=1$ by claim 2 of Properties of Natural Number Powers in a Field; and $2-\tfrac{1}{1}=1$, so the required inequality is $1\le1$.

Next, let $M\in P$. By claim 1 of Arithmetic of Addition on the Natural Numbers the successor of $M$ is $M+1$, so by the recursion in claim 1 of Properties of Finite Sums,

$$\sum_{m=1}^{M+1}\frac{1}{m^{2}}=\sum_{m=1}^{M}\frac{1}{m^{2}}+\frac{1}{(M+1)^{2}} .$$

The real numbers $M$ and $M+1$ are positive, so $M(M+1)$ and $(M+1)^{2}$ are positive by claim 5 of Elementary Order Arithmetic in an Ordered Field. From $M\le M+1$ and claim 5 of Elementary Arithmetic in an Ordered Field, with the nonnegative multiplier $M+1$, we get $M(M+1)\le(M+1)^{2}$, whence by claim 1 above

$$\frac{1}{(M+1)^{2}}\le\frac{1}{M(M+1)} .$$

Since $M$ and $M+1$ are nonzero, field arithmetic gives $\tfrac{1}{M}-\tfrac{1}{M+1}=\tfrac{1}{M(M+1)}$. Adding $\sum_{m=1}^{M}\tfrac{1}{m^{2}}\le2-\tfrac{1}{M}$ to the displayed inequality and using the identity just recorded,

$$\sum_{m=1}^{M+1}\frac{1}{m^{2}}\le 2-\frac{1}{M}+\frac{1}{M(M+1)}=2-\frac{1}{M+1},$$

so $M+1\in P$. Hence $P=\mathbb{N}$, which is clause 2.

Claim 3 (Clause 3). For every $m\in\mathbb{N}$ the real number $m^{2}$ is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field, so $\tfrac{1}{m^{2}}$ exists and is positive by claim 7 of that lemma; in particular $0\le\tfrac{1}{m^{2}}$. Let $s_{M}=\sum_{m=1}^{M}\tfrac{1}{m^{2}}$ denote the partial sums. Also $0\le\tfrac{1}{M}$ for every $M\in\mathbb{N}$, so $2-\tfrac{1}{M}\le2$; with clause 2 and transitivity, $s_{M}\le2$ for every $M\in\mathbb{N}$. Thus $\{s_{M}:M\in\mathbb{N}\}$ is bounded above by $2$, so by claim 1 of Series of Nonnegative Real Numbers, Comparison, and the Geometric Series the series converges with sum $\sup\{s_{M}:M\in\mathbb{N}\}$; and that least upper bound is at most $2$, since $2$ is an upper bound of the set.

Claim 4 (Denominators in clause 4). Write $N=M+M+1$ and let $c:[N]\to\mathbb{R}$ be the map with

$$c_{m}=\frac{1}{1+\alpha\,(m-M-1)^{2}} .$$

This is defined: $(m-M-1)^{2}$ is nonnegative by claim 2 of Nonnegativity of Squares in an Ordered Field, so $\alpha(m-M-1)^{2}$ is nonnegative by claim 5 of Elementary Arithmetic in an Ordered Field applied to $0\le(m-M-1)^{2}$ with the nonnegative multiplier $\alpha$, together with $\alpha\cdot0=0$ from claim 1 of Zero Products and Elementary Identities in a Field; hence $1\le1+\alpha(m-M-1)^{2}$, and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, so the denominator is positive by the mixed transitivity of claim 2 of that lemma and its inverse exists by claim 7.

Claim 5 (Splitting the sum of clause 4). By the associativity and commutativity of addition on $\mathbb{N}$, claims 3 and 4 of Arithmetic of Addition on the Natural Numbers, one has $N=M+(M+1)$ and $M+1=1+M$. Applying Splitting a Finite Sum at an Index to $c$ with the decomposition $N=M+(M+1)$ gives

$$\sum_{m=1}^{N}c_{m}=\sum_{m=1}^{M}c_{m}+\sum_{j=1}^{M+1}c_{M+j},$$

and applying it again to the map $j\mapsto c_{M+j}$ on $[1+M]$ with the decomposition $M+1=1+M$ gives

$$\sum_{j=1}^{M+1}c_{M+j}=c_{M+1}+\sum_{r=1}^{M}c_{M+1+r},$$

the first summand being a sum with a single term, evaluated by claim 1 of Properties of Finite Sums.

Claim 6 (The three pieces). For $r\in[M]$ one has $(M+1+r)-M-1=r$, so

$$c_{M+1+r}=\frac{1}{1+\alpha r^{2}} ;$$

and $(M+1)-M-1=0$, so $c_{M+1}=\tfrac{1}{1+\alpha\cdot0}=1$, using claim 1 of Zero Products and Elementary Identities in a Field.

For the first piece, define $\rho:[M]\to[M]$ as follows. Let $r\in[M]$, so that $r\le M$, and let $M<M+1$ be as given by claim 6 of Properties of the Order on the Natural Numbers. Then $r<M+1$. Indeed, by the trichotomy of claim 3 of that lemma exactly one of $r<M$, $r=M$, $M<r$ holds, and the last is impossible, since $M<r$ would give $M\le r$ by claim 1 and hence $r=M$ by the antisymmetry of claim 2, contradicting that exclusivity; in the case $r=M$ one has $r<M+1$ at once, and in the case $r<M$ the strict transitivity of claim 1 gives $r<M+1$. By claim 7 of that lemma there is exactly one $k\in\mathbb{N}$ with $M+1=r+k$, and we set $\rho(r)=k$. This $k$ lies in $[M]$: by claim 4 of Arithmetic of Addition on the Natural Numbers one has $r+k=k+r$, and $k<k+r$ by claim 6 of Properties of the Order on the Natural Numbers, so $k<M+1$, and then claim 5 of that lemma, applied with $M+1$ the successor of $M$, gives $k\le M$. The map $\rho$ is injective, since $\rho(r)=\rho(r')$ gives $r+\rho(r)=M+1=r'+\rho(r)$ and hence $r=r'$ by the cancellation of claim 5 of Arithmetic of Addition on the Natural Numbers together with claim 4 of that lemma; so $\rho$ is a bijection of $[M]$ onto itself by claim 1 of An Injective Self-Map of a Finite Set is a Bijection.

Read in $\mathbb{R}$, the identity $M+1=r+\rho(r)$ gives $\rho(r)-M-1=-r$, by the additivity of claim 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; hence $(\rho(r)-M-1)^{2}=r^{2}$ by claim 2 of Zero Products and Elementary Identities in a Field, and so $c_{\rho(r)}=\tfrac{1}{1+\alpha r^{2}}$. By claim 1 of Invariance of Finite Sums and Products under Reindexing by a Permutation, applied to the restriction of $c$ to $[M]$ and to $\rho$,

$$\sum_{r=1}^{M}c_{r}=\sum_{r=1}^{M}c_{\rho(r)}=\sum_{r=1}^{M}\frac{1}{1+\alpha r^{2}} .$$

Claim 7 (Clause 4). Let $r\in[M]$. The real number $\alpha r^{2}$ is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field, and $\alpha r^{2}\le1+\alpha r^{2}$ because $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field; so claim 1 above gives

$$\frac{1}{1+\alpha r^{2}}\le\frac{1}{\alpha r^{2}}=\frac{1}{\alpha}\cdot\frac{1}{r^{2}} ,$$

the last equality being field arithmetic with the nonzero factors $\alpha$ and $r^{2}$. Summing these inequalities over $r\in[M]$ by claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, and using the homogeneity of claim 3 of Properties of Finite Sums,

$$\sum_{r=1}^{M}\frac{1}{1+\alpha r^{2}}\le\frac{1}{\alpha}\sum_{r=1}^{M}\frac{1}{r^{2}} .$$

By claim 3 the sum on the right is at most $2$ — indeed $\sum_{r=1}^{M}\tfrac{1}{r^{2}}\le2$ was shown there — and $\tfrac{1}{\alpha}$ is nonnegative by claim 7 of Elementary Order Arithmetic in an Ordered Field, so claim 5 of Elementary Arithmetic in an Ordered Field gives

$$\sum_{r=1}^{M}\frac{1}{1+\alpha r^{2}}\le\frac{2}{\alpha} .$$

By claims 5 and 6 the sum of clause 4 equals $\sum_{r=1}^{M}c_{r}+1+\sum_{r=1}^{M}c_{M+1+r}$, and both of these sums equal $\sum_{r=1}^{M}\tfrac{1}{1+\alpha r^{2}}$. Adding the two bounds just obtained to $1\le1$ therefore gives

$$\sum_{m=1}^{2M+1}\frac{1}{1+\alpha\,(m-M-1)^{2}}\le\frac{2}{\alpha}+1+\frac{2}{\alpha}=1+\frac{4}{\alpha},$$

which is clause 4.