Proof of The Harmonic Series Diverges

Each result cited below is universally quantified over the data in its own statement.

Conventions. Let $\iota:\mathbb{N}\to\mathbb{R}$ be the canonical map through which natural numbers are read in $\mathbb{R}$ (The Real Numbers: Standing Notation and Background §numbers). By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, $0<\iota(m)$ and $\iota(m)^{-1}$ exists with $0<\iota(m)^{-1}$ for every $m\in\mathbb{N}$, so $h(m)=\tfrac1m=\iota(m)^{-1}$ defines a sequence $h:\mathbb{N}\to\mathbb{R}$ of positive terms. For $M\in\mathbb{N}$ let

$$H(M)=\sum_{m=1}^{M}\frac1m ,$$

the finite sum of the restriction of $h$ to $[M]$ (The Real Numbers: Standing Notation and Background §naturals); these are the partial sums of the series $\sum_{m=1}^{\infty}\tfrac1m$ (Series of Real Numbers §partial-sums). The order of $\mathbb{R}$ is a total order, reflexive, antisymmetric and transitive by clauses 1, 2 and 3 of Total Order on a Set; $x<y$ means $x\le y$ and $x\ne y$. Put $2=1+1$; by claim 8 of Elementary Order Arithmetic in an Ordered Field, $0<2$ and the inverse $2^{-1}$ exists, and $0<2^{-1}$ by claim 7 of that lemma.

Step 1 (recursion). $H(1)=1$, and $H(S(M))=H(M)+h(S(M))$ for every $M\in\mathbb{N}$. Indeed, by claim 1 of Properties of Finite Sums, $H(1)=h(1)=\iota(1)^{-1}$, and $\iota(1)=1$ by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, so $H(1)=1^{-1}=1$ by the field axioms of Field. For the second identity apply claim 1 of Properties of Finite Sums to the restriction of $h$ to $[S(M)]$; this is legitimate since $M<S(M)$ by claim 5 of Properties of the Order on the Natural Numbers, hence $M\le S(M)$ and $S(M)\in[S(M)]$ by claim 1 of Properties of the Order on the Natural Numbers, so $[M]\subseteq[S(M)]$ by transitivity (claim 1 of Properties of the Order on the Natural Numbers), and by claim 1 of Properties of Finite Sums the sum up to $M$ does not depend on which extension of the summands is used.

Step 2 (reciprocals reverse the order). If $x,y\in\mathbb{R}$ and $0<x\le y$, then $y^{-1}\le x^{-1}$. By mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) $0<y$, so $x^{-1}$ and $y^{-1}$ exist and are positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and $0<x^{-1}y^{-1}$ by claim 5 of Elementary Order Arithmetic in an Ordered Field. Multiplying $x\le y$ by the nonnegative number $x^{-1}y^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $x^{-1}y^{-1}x\le x^{-1}y^{-1}y$, which by the field axioms of Field reads $y^{-1}\le x^{-1}$. Consequently, for $m,m'\in\mathbb{N}$ with $m<m'$ we have $\iota(m)<\iota(m')$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, and with $0<\iota(m)$ this gives $h(m')\le h(m)$.

Step 3 (a block estimate). For every $p\in\mathbb{N}$ and every $M\in\mathbb{N}$,

$$H(M)+\iota(p)\,h(M+p)\le H(M+p).$$

Let $A$ be the set of those $p\in\mathbb{N}$ for which this holds for every $M\in\mathbb{N}$; we show $A=\mathbb{N}$ by the principle of induction of Principle of Induction for the Natural Numbers, applied to the set $A$.

(a) $1\in A$. Let $M\in\mathbb{N}$. By clause 1 of Natural Numbers, $M+1=S(M)$, and $\iota(1)=1$ by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; so the required inequality reads $H(M)+h(S(M))\le H(S(M))$, which holds with equality by Step 1, hence by reflexivity.

(b) If $p\in A$ then $S(p)\in A$. Let $M\in\mathbb{N}$ and put $q=M+p$. By clause 2 of Natural Numbers, $M+S(p)=S(q)$. Since $q<S(q)$ by claim 5 of Properties of the Order on the Natural Numbers, Step 2 gives $h(S(q))\le h(q)$; as $0\le\iota(p)$ (claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field), claim 5 of Elementary Arithmetic in an Ordered Field gives $\iota(p)h(S(q))\le\iota(p)h(q)$. Adding $H(M)$ (clause 1 of Ordered Field) and using $p\in A$ and transitivity,

$$H(M)+\iota(p)h(S(q))\le H(M)+\iota(p)h(q)\le H(q).$$

Adding $h(S(q))$ (clause 1 of Ordered Field) and using Step 1,

$$H(M)+\iota(p)h(S(q))+h(S(q))\le H(q)+h(S(q))=H(S(q)).$$

By distributivity (Field) the left-hand side is $H(M)+\bigl(\iota(p)+1\bigr)h(S(q))$, and $\iota(p)+1=\iota(p+1)=\iota(S(p))$ by claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and clause 1 of Natural Numbers. Since $S(q)=M+S(p)$, this is the required inequality for $S(p)$ and $M$. As $M$ was arbitrary, $S(p)\in A$.

By Principle of Induction for the Natural Numbers, $A=\mathbb{N}$.

Step 4 (doubling adds at least one half). For every $M\in\mathbb{N}$, $H(M)+2^{-1}\le H(M+M)$. Take $p=M$ in Step 3. By claim 4 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and the field axioms of Field, $\iota(M+M)=\iota(M)+\iota(M)=\iota(M)\cdot2$, and since $\iota(M)\ne0$ and $2\ne0$ the field axioms give $h(M+M)=(\iota(M)\cdot2)^{-1}=\iota(M)^{-1}2^{-1}$, so $\iota(M)h(M+M)=2^{-1}$. Step 3 now reads $H(M)+2^{-1}\le H(M+M)$.

Step 5 (growth). For every $j\in\mathbb{N}$ there is $M\in\mathbb{N}$ with $\iota(j)2^{-1}\le H(M)$. Let $A'$ be the set of those $j\in\mathbb{N}$ for which such an $M$ exists; we apply Principle of Induction for the Natural Numbers to $A'$.

(a) $1\in A'$: since $0<1$ (claim 6 of Elementary Order Arithmetic in an Ordered Field), claim 8 of Elementary Order Arithmetic in an Ordered Field gives $1\cdot2^{-1}<1$, and $\iota(1)=1$ (claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field) and $H(1)=1$ (Step 1), so $\iota(1)2^{-1}\le H(1)$.

(b) If $j\in A'$, witnessed by $M$, then by clause 1 of Natural Numbers, claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and distributivity, $\iota(S(j))2^{-1}=\iota(j)2^{-1}+2^{-1}$. Adding $2^{-1}$ to $\iota(j)2^{-1}\le H(M)$ (clause 1 of Ordered Field) and applying Step 4 and transitivity,

$$\iota(S(j))2^{-1}\le H(M)+2^{-1}\le H(M+M),$$

so $S(j)\in A'$, witnessed by $M+M$.

By Principle of Induction for the Natural Numbers, $A'=\mathbb{N}$.

Step 6 (the partial sums are not bounded above). Suppose $b\in\mathbb{R}$ were an upper bound of $\{H(M):M\in\mathbb{N}\}$ in the sense of The Real Numbers: Standing Notation and Background §bounds. Since $0<2^{-1}$, claim 2 of The Archimedean Property of the Real Numbers gives $j\in\mathbb{N}$ with $b<\iota(j)2^{-1}$, and Step 5 gives $M$ with $\iota(j)2^{-1}\le H(M)$. By mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field), $b<H(M)$. But $H(M)\le b$, so antisymmetry gives $b=H(M)$, contradicting $b\ne H(M)$. Hence the set is not bounded above.

Step 7 (divergence). The terms $h(m)$ are nonnegative, being positive. By Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion the series $\sum_{m=1}^{\infty}\tfrac1m$ converges if and only if the set of its partial sums $\{H(M):M\in\mathbb{N}\}$ is bounded above, which fails by Step 6. So the series does not converge in the sense of Series of Real Numbers §convergent.