Proof of Arithmetic of Limits of Real Sequences

For a real number $x$, $|x|$ denotes its absolute value, which by claim 8 of Properties of Complex Conjugation and Modulus is the modulus of $x$ as a complex number; thus $0\le|x|$, $|x|=0$ only for $x=0$ (claim 3), $|xy|=|x|\,|y|$ (claim 4) and $|x+y|\le|x|+|y|$ (claim 7), and $|-x|=|x|$ because $|-1|=1$. Convergence is as in Limit of a Sequence of Real Numbers. We use the ordered field properties of $\mathbb{R}$, in particular that if $0<p\le q$ then $1/q\le1/p$, which follows by multiplying $p\le q$ by the positive number $1/(pq)$.

Claim 1. Let $\varepsilon$ be a positive real number. Choose $N_{1}$ with $|a_{n}-A|<\varepsilon/2$ for $n\ge N_{1}$ and $N_{2}$ with $|b_{n}-B|<\varepsilon/2$ for $n\ge N_{2}$, and let $N$ be the larger. For $n\ge N$,

Claim 2. By claim 2 of Uniqueness of Limits and Boundedness of Convergent Real Sequences there is a positive real $M$ with $|a_{n}|\le M$ for every $n$. Put $K=M+|B|+1$, so that $K$ is positive and $M+|B|\le K$. Let $\varepsilon$ be positive; then $\varepsilon/(2K)$ is positive. Choose $N_{1}$ with $|a_{n}-A|<\varepsilon/(2K)$ for $n\ge N_{1}$, and $N_{2}$ with $|b_{n}-B|<\varepsilon/(2K)$ for $n\ge N_{2}$; let $N$ be the larger. For $n\ge N$, using $a_{n}b_{n}-AB=a_{n}(b_{n}-B)+(a_{n}-A)B$,

the strict inequality in the middle because $M$ is positive. Hence $(a_{n}b_{n})$ converges to $AB$.

Claim 3. The constant sequence with every term equal to $c$ converges to $c$, since $|c-c|=0<\varepsilon$ for every positive $\varepsilon$. Applying claim 2 to $(a_{n})$ and this constant sequence gives that $(a_{n}c)=(ca_{n})$ converges to $Ac=cA$. Taking $c=-1$ shows $(-a_{n})$ converges to $-A$, and then claim 1 applied to $(a_{n})$ and $(-b_{n})$ shows that $(a_{n}-b_{n})$ converges to $A-B$.

Claim 4. We first show that $(1/b_{n})$ converges to $1/B$. Since $B\neq0$ we have $0<|B|$, so $|B|/2$ is positive. Choose $N_{0}$ with $|b_{n}-B|<|B|/2$ for $n\ge N_{0}$. For such $n$,

so $|B|/2<|b_{n}|$, and in particular $1/|b_{n}|<2/|B|$. Hence, for $n\ge N_{0}$,

Let $\varepsilon$ be positive; then $\varepsilon|B|^{2}/2$ is positive, so there is $N_{1}$ with $|b_{n}-B|<\varepsilon|B|^{2}/2$ for $n\ge N_{1}$. For $n$ at least as large as both $N_{0}$ and $N_{1}$ we get $|1/b_{n}-1/B|<\varepsilon$. Thus $(1/b_{n})$ converges to $1/B$, and applying claim 2 to $(a_{n})$ and $(1/b_{n})$ shows that $(a_{n}/b_{n})=(a_{n}\cdot(1/b_{n}))$ converges to $A\cdot(1/B)=A/B$.