Proof of Finite Sums of Real Numbers: Nonnegativity, Domination by the Sum, and Limits

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement above. We use silently that the order of $\mathbb{R}$ is reflexive and transitive and that equal real numbers satisfy $\le$ in both directions. Let $S$ denote the successor map of Natural Numbers.

Two facts about initial segments. First, if $S(j)\in[N]$ for a natural number $j$, then $j\in[N]$: indeed $S(j)\le N$ by Initial Segment of the Natural Numbers, while $j<S(j)$ and hence $j\le S(j)$ by claims 5 and 1 of Properties of the Order on the Natural Numbers, so $j\le N$ by claim 1 of that lemma and $j\in[N]$ by Initial Segment of the Natural Numbers. Second, $[1]=\{1\}$: if $k\in[1]$ then $k\le 1$, while $1\le k$ by claim 4 of Properties of the Order on the Natural Numbers, so $k=1$ by claim 2 of that lemma.

The base and recursion identities. Let $a:[N]\to\mathbb{R}$ be a map. By Finite Sum Notation in a Field,

We refer to these below as the base and the recursion identity for $a$. Each of the three claims is proved by induction on a natural number $j$, the assertion being formulated so as to be vacuously true when $j\notin[N]$; since the claim supplies $j\in[N]$, this suffices.

Claim 1. Let $t:[N]\to\mathbb{R}$ satisfy $0\le t_{k}$ for every $k\in[N]$. For a natural number $j$ let $P(j)$ assert: if $j\in[N]$ then $0\le\sum_{k=1}^{j}t_{k}$.

If $1\in[N]$ then $\sum_{k=1}^{1}t_{k}=t_{1}$ by the base identity, and $0\le t_{1}$ by hypothesis; so $P(1)$ holds. Assume $P(j)$ and suppose $S(j)\in[N]$. Then $j\in[N]$ by the first fact above, so $0\le\sum_{k=1}^{j}t_{k}$ by $P(j)$, and $0\le t_{S(j)}$ by hypothesis. By the recursion identity and claim 2 of Elementary Arithmetic in an Ordered Field, $0\le\sum_{k=1}^{S(j)}t_{k}$. Hence $P(S(j))$, and the induction is complete.

For the second assertion of the claim, suppose $t_{k}=0$ for every $k\in[N]$ and let $P'(j)$ assert: if $j\in[N]$ then $\sum_{k=1}^{j}t_{k}=0$. If $1\in[N]$ then $\sum_{k=1}^{1}t_{k}=t_{1}=0$ by the base identity. Assume $P'(j)$ and suppose $S(j)\in[N]$; then $j\in[N]$, and by the recursion identity $\sum_{k=1}^{S(j)}t_{k}=0+t_{S(j)}=0+0=0$. Hence $P'(S(j))$.

Claim 2. Let $t$ be as in claim 1. For a natural number $j$ let $R(j)$ assert: if $j\in[N]$ then $t_{i}\le\sum_{k=1}^{j}t_{k}$ for every $i\in[j]$.

If $1\in[N]$, then by the second fact above the only $i\in[1]$ is $1$, and $\sum_{k=1}^{1}t_{k}=t_{1}$ by the base identity; so $R(1)$ holds.

Assume $R(j)$ and suppose $S(j)\in[N]$, so that $j\in[N]$. Let $i\in[S(j)]$, and write $\Sigma_{j}=\sum_{k=1}^{j}t_{k}$ and $\Sigma_{S(j)}=\sum_{k=1}^{S(j)}t_{k}$, so that $\Sigma_{S(j)}=\Sigma_{j}+t_{S(j)}$ by the recursion identity.

If $i=S(j)$, then $\Sigma_{S(j)}-t_{i}=\Sigma_{j}$, which is nonnegative by claim 1; hence $t_{i}\le\Sigma_{S(j)}$ by claim 3 of Elementary Arithmetic in an Ordered Field.

If $i\ne S(j)$, then $i\le j$ by claim 5 of Properties of the Order on the Natural Numbers, so $i\in[j]$ by Initial Segment of the Natural Numbers and $t_{i}\le\Sigma_{j}$ by $R(j)$. Moreover $\Sigma_{S(j)}-\Sigma_{j}=t_{S(j)}$, which is nonnegative by hypothesis, so $\Sigma_{j}\le\Sigma_{S(j)}$ by claim 3 of Elementary Arithmetic in an Ordered Field. By transitivity $t_{i}\le\Sigma_{S(j)}$.

In both cases $t_{i}\le\Sigma_{S(j)}$, so $R(S(j))$ holds and the induction is complete.

Claim 3. For a natural number $j$ let $L(j)$ assert: if $j\in[N]$ then the sequence whose $m$th term is $\sum_{k=1}^{j}a_{k,m}$ converges to $\sum_{k=1}^{j}A_{k}$. Here, for each fixed $m$, the sum $\sum_{k=1}^{j}a_{k,m}$ is formed from the map $k\mapsto a_{k,m}$ on $[N]$, and $\sum_{k=1}^{j}A_{k}$ from the map $k\mapsto A_{k}$.

If $1\in[N]$, the base identity gives $\sum_{k=1}^{1}a_{k,m}=a_{1,m}$ for every $m$ and $\sum_{k=1}^{1}A_{k}=A_{1}$, and $(a_{1,m})_{m\in\mathbb{N}}$ converges to $A_{1}$ by hypothesis; so $L(1)$ holds.

Assume $L(j)$ and suppose $S(j)\in[N]$, so that $j\in[N]$. By the recursion identity, the $m$th term of the sequence associated with $S(j)$ is $\bigl(\sum_{k=1}^{j}a_{k,m}\bigr)+a_{S(j),m}$. By $L(j)$ the sequence whose $m$th term is $\sum_{k=1}^{j}a_{k,m}$ converges to $\sum_{k=1}^{j}A_{k}$, and by hypothesis $(a_{S(j),m})_{m\in\mathbb{N}}$ converges to $A_{S(j)}$. By claim 1 of Arithmetic of Limits of Real Sequences the sequence of sums converges to $\bigl(\sum_{k=1}^{j}A_{k}\bigr)+A_{S(j)}$, which equals $\sum_{k=1}^{S(j)}A_{k}$ by the recursion identity. Hence $L(S(j))$, and the induction is complete.