Proof of Series of Nonnegative Real Numbers, Comparison, and the Geometric Series

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 $s_{n+1}=s_{n}+a_{n+1}$.

Claim 1. Assume $0\le a_{k}$ for every $k$. From $s_{n+1}=s_{n}+a_{n+1}$ and claim 3 of Elementary Arithmetic in an Ordered Field we get $s_{n}\le s_{n+1}$ for every $n$. By induction on $n$ it follows that $s_{m}\le s_{n}$ whenever $m\le n$: the case $n=m$ is claim 1 of Properties of the Order on the Natural Numbers together with reflexivity of $\le$ on $\mathbb{R}$, and the inductive step is $s_{m}\le s_{n}\le s_{n+1}$. Also $0\le s_{1}=a_{1}$, so $0\le s_{n}$ for every $n$.

Let $P=\{s_{n}:n\in\mathbb{N}\}$, a nonempty subset of $\mathbb{R}$. If $\sum_{k=1}^{\infty}a_{k}$ converges, then $(s_{n})$ converges and is therefore bounded by claim 2 of Uniqueness of Limits and Boundedness of Convergent Real Sequences; in particular $P$ is bounded above.

Conversely, suppose $P$ is bounded above and let $L=\sup P$, which exists and is unique by the least upper bound property. Let $\varepsilon$ be positive. By claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is $s_{N}\in P$ with $L-\varepsilon<s_{N}$. For $n\in\mathbb{N}$ with $N\le n$ we have $s_{N}\le s_{n}\le L$, the last because $L$ is an upper bound for $P$; hence $-\varepsilon<s_{n}-L\le0<\varepsilon$ and so $|s_{n}-L|<\varepsilon$ by claim 9 of Properties of the Absolute Value in an Ordered Field. Thus $(s_{n})$ converges to $L$, so the series converges with sum $L=\sup P$.

Claim 2. Under the hypotheses, claim 1 gives $\sum_{k=1}^{\infty}a_{k}=\sup P$, which is an upper bound for $P$, so $s_{n}\le\sum_{k=1}^{\infty}a_{k}$ for every $n$; and $0\le s_{n}$ was shown in claim 1.

Claim 3. By claims 2, 3 and 5 of Properties of Finite Sums, the $n$-th partial sum of the sequence with terms $b_{k}+(-1)a_{k}$, which are nonnegative, equals $t_{n}-s_{n}$ and is nonnegative; hence $s_{n}\le t_{n}$ for every $n$. Writing $T=\sum_{k=1}^{\infty}b_{k}$, claim 2 applied to $(b_{k})$ gives $t_{n}\le T$, so $s_{n}\le T$ for every $n$ and the set $\{s_{n}:n\in\mathbb{N}\}$ is bounded above. By claim 1 the series $\sum_{k=1}^{\infty}a_{k}$ converges, with sum $\sup\{s_{n}:n\in\mathbb{N}\}$; since $T$ is an upper bound for that set, $\sum_{k=1}^{\infty}a_{k}\le T$.

Claim 4. Since $r<1$, claim 1 of Elementary Order Arithmetic in an Ordered Field gives $0<1-r$, so $1-r$ is nonzero and has a positive multiplicative inverse by claim 7 of that lemma.

The sequence $(r^{k})$ converges to $0$. By claim 1 of Properties of Natural Number Powers in a Field, $r^{k+1}=r^{k}r$, and by claim 5 of that lemma $0\le r^{k}$. Hence $r^{k+1}=r^{k}r\le r^{k}\cdot1=r^{k}$ by claim 5 of Elementary Arithmetic in an Ordered Field, and by induction $r^{n}\le r^{m}$ whenever $m\le n$. The set $\{r^{k}:k\in\mathbb{N}\}$ is nonempty and bounded below by $0$, so $L=\inf\{r^{k}:k\in\mathbb{N}\}$ exists by the greatest lower bound property. Given a positive $\varepsilon$, claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ provides $N$ with $r^{N}<L+\varepsilon$, and then $L\le r^{k}\le r^{N}<L+\varepsilon$ for $N\le k$, so $|r^{k}-L|<\varepsilon$ by claim 9 of Properties of the Absolute Value in an Ordered Field. Hence $(r^{k})$ converges to $L$. The sequence $(r^{k+1})_{k\in\mathbb{N}}$ converges to $L$ as well, since $N\le k$ implies $N\le k+1$ by claims 1 and 5 of Properties of the Order on the Natural Numbers; on the other hand $(r^{k+1})=(r^{k}r)$ converges to $Lr$ by claim 3 of Arithmetic of Limits of Real Sequences. By uniqueness of limits, $L=Lr$, hence $L(1-r)=0$, and multiplying by $(1-r)^{-1}$ gives $L=0$.

The partial sums. We prove $\sum_{k=1}^{n}r^{k}=\frac{r-r^{n+1}}{1-r}$ by induction on $n$. For $n=1$: $(1-r)r=r-r^{2}$, so $r=\frac{r-r^{2}}{1-r}$. Assuming the identity for $n$,

using $r^{n+1}r=r^{n+2}$ from claim 1 of Properties of Natural Number Powers in a Field.

The sum and the tails. The sequence $(r^{n+1})_{n\in\mathbb{N}}$ converges to $0$, as shown above, so by claims 1 and 3 of Arithmetic of Limits of Real Sequences the sequence of partial sums converges to $\frac{r-0}{1-r}=\frac{r}{1-r}$. Hence $\sum_{k=1}^{\infty}r^{k}=\frac{r}{1-r}$, and subtracting the closed form of the $n$-th partial sum gives

For $r=\tfrac{1}{2}$ we have $1-r=\tfrac{1}{2}$, so $\frac{r}{1-r}=1$ and $\frac{r^{n+1}}{1-r}=2\,r^{n+1}=2\,r^{n}r=r^{n}$, since $2r=1$.

Claim 5. From $0\le\mu_{k}$ and $0\le w_{k}\le M$, claim 5 of Elementary Arithmetic in an Ordered Field gives $0\le\mu_{k}w_{k}$ and $\mu_{k}w_{k}\le\mu_{k}M$. By claim 1 of Elementary Properties of Series of Real Numbers the series with terms $\mu_{k}M$ converges, with sum $M\sum_{k=1}^{\infty}\mu_{k}$, so $\sum_{k=1}^{\infty}\mu_{k}w_{k}$ converges by claim 3 above. The left-hand inequality is claim 2 above applied to the sequence with terms $\mu_{k}w_{k}$.

For the right-hand inequality, put $v_{k}=\mu_{k}(M-w_{k})$. Then $0\le M-w_{k}$ and hence $0\le v_{k}$ by claim 5 of Elementary Arithmetic in an Ordered Field, and $v_{k}=\mu_{k}M-\mu_{k}w_{k}$. By claim 1 of Elementary Properties of Series of Real Numbers the series $\sum_{k=1}^{\infty}v_{k}$ converges with sum $M\sum_{k=1}^{\infty}\mu_{k}-\sum_{k=1}^{\infty}\mu_{k}w_{k}$, and by claims 2 and 3 of Properties of Finite Sums its $n$-th partial sum equals $M\sum_{k=1}^{n}\mu_{k}-\sum_{k=1}^{n}\mu_{k}w_{k}$. Claim 2 above, applied to $(v_{k})$, gives

and rearranging, by claim 3 of Elementary Arithmetic in an Ordered Field, yields the asserted bound.