Proof of Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality

Each result cited below is universally quantified over the data in its own statement and is applied to the data indicated where it is cited.

In a commutative ring the ring laws make $+$ associative and commutative, so the clauses of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms for associative and commutative operations apply to finite sums. Each induction below runs over the set of those $n\in\mathbb{N}$ for which the claim holds for all data of the stated kind, and concludes by induction from $1$ on $\mathbb{N}$, The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction. In the inductive steps, $\sum_{k=1}^{n}$ of a map on $[n+1]$ is the sum of its restriction to $[n]$ by Iterated Operations: Finite Sums and Finite Products §restriction, and the step uses Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion:

$$\sum_{k=1}^{1}a_{k}=a_{1},\qquad\sum_{k=1}^{n+1}a_{k}=\sum_{k=1}^{n}a_{k}+a_{n+1}.$$

Distributive. The map $x\mapsto c\,x$ on $R$ satisfies $c(x+y)=c\,x+c\,y$ by the ring laws, so the claim is Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §homomorphism with $\ast=\diamond=+$.

Difference. For $x,y\in R$, the ring laws give $(x+y)+\big((-x)+(-y)\big)=\big(x+(-x)\big)+\big(y+(-y)\big)=0$, so $-(x+y)=(-x)+(-y)$ by the uniqueness in Additive and Multiplicative Inverses Are Unique §negative. Thus $x\mapsto-x$ satisfies the hypothesis of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §homomorphism, which gives the first identity. For the second, Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §termwise and the first identity give

$$\sum_{k=1}^{n}\big(a_{k}+(-b_{k})\big)=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}(-b_{k})=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}.$$

Telescoping. We induct on $n$. For $n=1$ the sum is $d_{2}-d_{1}$. If the claim holds for $n$ and $d:[n+2]\to R$, then the induction hypothesis for $d|_{[n+1]}$ and the ring laws give

$$\sum_{k=1}^{n+1}(d_{k+1}-d_{k})=(d_{n+1}-d_{1})+(d_{n+2}-d_{n+1})=d_{n+2}-d_{1}.$$

Constant. Here $c\in R$, and by Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals the factor $n$ stands for its image $n_{R}$ in $R$. By the same clause, $1_{R}$ is the unit $1$ of $R$ and the image of a sum is the sum of the images, so $(n+1)_{R}=n_{R}+1_{R}=n_{R}+1$ for every $n\in\mathbb{N}$. We induct on $n$. For $n=1$ the sum is $c=1\cdot c=1_{R}\,c$. If $\sum_{k=1}^{n}c=n_{R}\,c$, then the recursion and the ring laws give

$$\sum_{k=1}^{n+1}c=n_{R}\,c+c=n_{R}\,c+1\cdot c=(n_{R}+1)\,c=(n+1)_{R}\,c.$$

From now on $F$ is an ordered field. Its underlying ring is a commutative ring, and the rules of Rules of Arithmetic and Order in an Ordered Field are available.

Comparison. We induct on $n$, proving both assertions together. For $n=1$ the sums are $a_{1}$ and $b_{1}$, and $j=1$. Suppose they hold for $n$, and let $a,b:[n+1]\to F$ with $a_{k}\le b_{k}$ for all $k$. Put $A=\sum_{k=1}^{n}a_{k}$ and $B=\sum_{k=1}^{n}b_{k}$. Then $A\le B$ by the induction hypothesis, so $A+a_{n+1}\le B+b_{n+1}$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, which is the first assertion. Now let $a_{j}<b_{j}$ with $j\in[n+1]$, so that $j\le n$ or $j=n+1$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §successor. If $j\le n$, then $A<B$ by the induction hypothesis, and Rules of Arithmetic and Order in an Ordered Field §order-sum gives $A+a_{n+1}<B+a_{n+1}\le B+b_{n+1}$. If $j=n+1$, it gives $A+a_{n+1}\le B+a_{n+1}=a_{n+1}+B<b_{n+1}+B$. In both cases $\sum_{k=1}^{n+1}a_{k}<\sum_{k=1}^{n+1}b_{k}$.

Nonnegative. We induct on $n$. For $n=1$, $j=1$ and $a_{1}=\sum_{k=1}^{1}a_{k}$. Suppose the claim holds for $n$, and let $a:[n+1]\to F$ with all $a_{k}\ge0$; put $A=\sum_{k=1}^{n}a_{k}$. By the induction hypothesis $0\le a_{1}\le A$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, $A=A+0\le A+a_{n+1}$ and $a_{n+1}=0+a_{n+1}\le A+a_{n+1}$. Hence for $j\le n$, $a_{j}\le A\le\sum_{k=1}^{n+1}a_{k}$ by the induction hypothesis, and for $j=n+1$, $a_{j}\le\sum_{k=1}^{n+1}a_{k}$ directly. In particular $0\le a_{1}\le\sum_{k=1}^{n}a_{k}$ for every $n$.

Triangle. We induct on $n$; for $n=1$ both sides are $|a_{1}|$. If the claim holds for $n$ and $a:[n+1]\to F$, then Rules of Arithmetic and Order in an Ordered Field §triangle, the induction hypothesis and Rules of Arithmetic and Order in an Ordered Field §order-sum give

$$\Big|\sum_{k=1}^{n+1}a_{k}\Big|\le\Big|\sum_{k=1}^{n}a_{k}\Big|+|a_{n+1}|\le\sum_{k=1}^{n}|a_{k}|+|a_{n+1}|=\sum_{k=1}^{n+1}|a_{k}|.$$