Proof of Elementary Properties of Series of Real Numbers

Throughout, $(s_{n})$ and $(t_{n})$ denote the partial sums of $(a_{k})$ and $(b_{k})$. We write $S$ for the successor map on $\mathbb{N}$ and $n+1$ for $S(n)$, and use the recursion of claim 1 of Properties of Finite Sums: $s_{1}=a_{1}$ and $s_{n+1}=s_{n}+a_{n+1}$ for every $n\in\mathbb{N}$.

An observation on indices. Every $k\in\mathbb{N}$ is either equal to $1$ or of the form $S(j)$ for some $j\in\mathbb{N}$: the set of those $k\in\mathbb{N}$ with this property contains $1$, and contains $S(m)$ whenever it contains $m$, hence is all of $\mathbb{N}$ by induction as in Natural Numbers.

Claim 1. By claim 2 of Properties of Finite Sums the $n$-th partial sum of $(a_{k}+b_{k})$ is $s_{n}+t_{n}$, and by claim 3 of that lemma the $n$-th partial sum of $(\lambda a_{k})$ is $\lambda s_{n}$. Suppose $(s_{n})$ converges to $L$ and $(t_{n})$ to $M$. Then $(s_{n}+t_{n})$ converges to $L+M$ by claim 1 of Arithmetic of Limits of Real Sequences, and $(\lambda s_{n})$ converges to $\lambda L$ by claim 3 of that theorem. By the definition of convergence of a series and of its sum, this is precisely the assertion.

Claim 2. Suppose $(s_{n})$ converges to $L$, and let $\varepsilon\in\mathbb{R}$ be positive; then $\tfrac{\varepsilon}{2}$ is positive and $\tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon$ by claim 8 of Elementary Order Arithmetic in an Ordered Field. Choose $N\in\mathbb{N}$ with $|s_{n}-L|<\tfrac{\varepsilon}{2}$ for every $n\in\mathbb{N}$ with $N\le n$, and put $N'=S(N)$. Let $k\in\mathbb{N}$ satisfy $N'\le k$. We check that $k\ne1$. Suppose $k=1$, so that $S(N)\le1$. By claim 4 of Properties of the Order on the Natural Numbers we have $1\le N$, and $N<S(N)$ by claim 5, hence $N\le S(N)$ by claim 1. Transitivity of $\le$, claim 1, then gives $N\le1$ and $S(N)\le N$, so $N=1$ and $N=S(N)$ by claim 2, whence $N<N$, which claim 2 excludes. Therefore $k\ne1$. By the observation above there is $j\in\mathbb{N}$ with $k=S(j)$, and the recursion gives $a_{k}=a_{S(j)}=s_{S(j)}-s_{j}=s_{k}-s_{j}$.

We check that $N\le j$. Suppose not; then $j<N$ by trichotomy, claim 3 of Properties of the Order on the Natural Numbers, hence $j\le N$ by claim 1 and $S(j)\le S(N)$ by claim 6 of that lemma. Since $j\ne N$ and $S$ is injective by Natural Numbers, we have $S(j)\ne S(N)$. But $S(N)=N'\le k=S(j)$, so $S(j)\le S(N)$ and $S(N)\le S(j)$ give $S(j)=S(N)$ by claim 2 of Properties of the Order on the Natural Numbers, a contradiction. Hence $N\le j$, and also $N\le k$ because $N\le N'\le k$ by claims 1 and 5.

Therefore, by claims 2 and 5 of Properties of the Absolute Value in an Ordered Field,

As $\varepsilon$ was an arbitrary positive real number, $(a_{k})_{k\in\mathbb{N}}$ converges to $0$.

Claim 3. Suppose first that $(s_{n})$ converges to $L$, and let $\varepsilon$ be positive. Choose $N$ with $|s_{n}-L|<\tfrac{\varepsilon}{2}$ for $N\le n$. For $m,n\in\mathbb{N}$ with $N\le m$ and $N\le n$, claims 2 and 5 of Properties of the Absolute Value in an Ordered Field give

Conversely, suppose the stated condition holds. Since $d_{\mathbb{R}}(s_{n},s_{m})=|s_{n}-s_{m}|$, it says exactly that $(s_{n})$ is a Cauchy sequence in $(\mathbb{R},d_{\mathbb{R}})$, hence a Cauchy sequence of real numbers, and therefore convergent by the completeness recorded there. In both directions the conclusion is the assertion about the series, by the definition of convergence of a series.

Claim 4. For every $k\in\mathbb{N}$ we have $0\le b_{k}-a_{k}$, by claim 3 of Elementary Arithmetic in an Ordered Field applied to $a_{k}\le b_{k}$. By claims 2 and 3 of Properties of Finite Sums, the $n$-th partial sum of the sequence with terms $b_{k}+(-1)a_{k}$ equals $t_{n}-s_{n}$; by claim 5 of that lemma it is nonnegative, so $s_{n}\le t_{n}$ for every $n\in\mathbb{N}$, again by claim 3 of Elementary Arithmetic in an Ordered Field. Claim 1 of Order Properties of Limits of Real Sequences, applied to the convergent sequences $(s_{n})$ and $(t_{n})$, gives the stated inequality between their limits.

Claim 5. We show $s_{n}=c_{n+1}-c_{1}$ by induction on $n$. For $n=1$ we have $s_{1}=a_{1}=c_{2}-c_{1}$. If $s_{n}=c_{n+1}-c_{1}$, then

which is the identity for $n+1$. Hence $s_{n}=c_{n+1}-c_{1}$ for every $n\in\mathbb{N}$.

It remains to compare the convergence of $(s_{n})$ with that of $(c_{k})_{k\in\mathbb{N}}$. Suppose $(c_{k})$ converges to $M$ and let $\varepsilon$ be positive; choose $N$ with $|c_{k}-M|<\varepsilon$ for $N\le k$. For $n\in\mathbb{N}$ with $N\le n$ we have $n<n+1$ by claim 5 of Properties of the Order on the Natural Numbers, hence $N\le n+1$ by claim 1, so $|c_{n+1}-M|<\varepsilon$. Thus $(c_{n+1})_{n\in\mathbb{N}}$ converges to $M$, and $(s_{n})$ converges to $M-c_{1}$ by claims 1 and 3 of Arithmetic of Limits of Real Sequences.

Conversely, suppose $(s_{n})$ converges, say to $P$; then $(c_{n+1})_{n\in\mathbb{N}}$ converges to $P+c_{1}=:M$ by the same two claims. Let $\varepsilon$ be positive and choose $N$ with $|c_{n+1}-M|<\varepsilon$ for $N\le n$; put $N'=S(N)$. For $k\in\mathbb{N}$ with $N'\le k$, the index argument used in claim 2 gives $j\in\mathbb{N}$ with $k=S(j)=j+1$ and $N\le j$, so $|c_{k}-M|=|c_{j+1}-M|<\varepsilon$. Hence $(c_{k})$ converges to $M$, and $\sum_{k=1}^{\infty}a_{k}=P=M-c_{1}$, as asserted.