Proof of Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series

Each result cited is universally quantified over the data in its own statement. Write $S=\{s_{n}:n\in\mathbb{N}\}$, $t_{n}=\sum_{k=1}^{n}b_{k}$ and $u_{n}=\sum_{k=1}^{n}|a_{k}|$ for $n\in\mathbb{N}$; then $(t_{n})$ and $(u_{n})$ are the sequences of partial sums of $(b_{k})$ and of $(|a_{k}|)$. By Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges, Convergent Sequences of Real Numbers §converges and The Limit of a Convergent Sequence §limit, a series converges if and only if its sequence of partial sums converges to some real number, and its sum is then that number.

Clause bounded. Let $a_{k}\ge0$ for every $k\in\mathbb{N}$. For $n\in\mathbb{N}$, $s_{n+1}=s_{n}+a_{n+1}$ by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §interval-recursion (with $m=1$), and $a_{n+1}\ge0$, so $s_{n}\le s_{n+1}$; thus $(s_{n})$ is nondecreasing by Monotone Sequences §monotone. Also $0\le s_{n}$ for every $n$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §nonnegative. If $(s_{n})$ is bounded above, then $s_{n}\to\sup S$ by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences §monotone, so by the opening paragraph the series $\sum_{k=1}^{\infty}a_{k}$ converges with sum $\sup S$. Conversely, if it converges, then $(s_{n})$ is convergent by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges, hence bounded by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §bounded, hence bounded above by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §two-sided, and its sum is $\sup S$ as just shown. In that case $\sup S$ is an upper bound of $S$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds, so $0\le s_{n}\le\sup S=\sum_{k=1}^{\infty}a_{k}$ for every $n\in\mathbb{N}$.

Clause comparison. Let $0\le a_{k}\le b_{k}$ for every $k\in\mathbb{N}$, and let $\sum_{k=1}^{\infty}b_{k}$ converge, with sum $B$. As $b_{k}\ge0$ for every $k$, clause bounded, applied to $(b_{k})$, gives $t_{n}\le B$ for every $n\in\mathbb{N}$. By Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison, $s_{n}\le t_{n}\le B$ for every $n$. Hence $B$ is an upper bound of $S$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds, so $S$ is bounded above by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded, that is, $(s_{n})$ is bounded above by Bounded Sequences of Real Numbers §bounded. By clause bounded, applied to $(a_{k})$, the series $\sum_{k=1}^{\infty}a_{k}$ converges; with $A$ its sum, $s_{n}\to A$, and since $s_{n}\le B$ for every $n\ge1$, $A\le B$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §order.

Now let $n\in\mathbb{N}$. As $\sum_{k=1}^{\infty}a_{k}$ and $\sum_{k=1}^{\infty}b_{k}$ converge, Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping §shift, applied with $p=n$ to $(a_{k})$ and to $(b_{k})$, shows that $\sum_{k=1}^{\infty}a_{n+k}$ and $\sum_{k=1}^{\infty}b_{n+k}$ converge. Since $a_{n+k}\le b_{n+k}$ for every $k\in\mathbb{N}$, Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping §order, applied to the sequences $(a_{n+k})_{k\in\mathbb{N}}$ and $(b_{n+k})_{k\in\mathbb{N}}$, gives $\sum_{k=1}^{\infty}a_{n+k}\le\sum_{k=1}^{\infty}b_{n+k}$. Since $a_{n+k}\ge0$ for every $k\in\mathbb{N}$, clause bounded, applied to $(a_{n+k})_{k\in\mathbb{N}}$, gives $0\le\sum_{j=1}^{1}a_{n+j}\le\sum_{k=1}^{\infty}a_{n+k}$, so $0\le\sum_{k=1}^{\infty}a_{n+k}$.

Clause absolute. Let $\sum_{k=1}^{\infty}a_{k}$ converge absolutely. By Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §absolute the series $\sum_{k=1}^{\infty}|a_{k}|$ converges; let $U$ be its sum, so $u_{n}\to U$. For $k\in\mathbb{N}$ put $p_{k}=a_{k}+|a_{k}|$; since $-|a_{k}|\le a_{k}\le|a_{k}|$, we have $0\le p_{k}\le2|a_{k}|$. The partial sums of $(2|a_{k}|)$ are $\sum_{k=1}^{n}2|a_{k}|=2u_{n}$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive, and $2u_{n}\to2U$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic, so $\sum_{k=1}^{\infty}2|a_{k}|$ converges. By clause comparison above, applied to $(p_{k})$ and $(2|a_{k}|)$, the series $\sum_{k=1}^{\infty}p_{k}$ converges; let $P$ be its sum, so $P_{n}=\sum_{k=1}^{n}p_{k}\to P$. As $a_{k}=p_{k}-|a_{k}|$ for every $k$, $s_{n}=P_{n}-u_{n}$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §difference, so $s_{n}\to P-U$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic. Hence $\sum_{k=1}^{\infty}a_{k}$ converges; let $A$ be its sum, so $s_{n}\to A$. Then $|s_{n}|\to|A|$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §absolute, and $|s_{n}|\le u_{n}$ for every $n\ge1$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §triangle, so $|A|\le U$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §order.

Clause dominated. Let $|a_{k}|\le b_{k}$ for every $k\in\mathbb{N}$, and let $\sum_{k=1}^{\infty}b_{k}$ converge. Then $0\le|a_{k}|\le b_{k}$ for every $k$, so by clause comparison above, applied to $(|a_{k}|)$ and $(b_{k})$, the series $\sum_{k=1}^{\infty}|a_{k}|$ converges and $\sum_{k=1}^{\infty}|a_{k}|\le\sum_{k=1}^{\infty}b_{k}$. Thus $\sum_{k=1}^{\infty}a_{k}$ converges absolutely, and by clause absolute above it converges with $\big|\sum_{k=1}^{\infty}a_{k}\big|\le\sum_{k=1}^{\infty}|a_{k}|$.

Clause geometric. First let $x\in\mathbb{R}$ with $|x|<1$. Then $x\le|x|<1$, so $x\neq1$ and $1-x\neq0$. For $n\in\mathbb{N}$, $\sum_{k=1}^{n}x^{k-1}=\frac{1-x^{n}}{1-x}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §geometric. For $k\in\mathbb{N}$, $1+(k-1)=k$ by The Difference of Two Natural Numbers with Zero §difference, so $x^{k}=x^{1}x^{k-1}=x\,x^{k-1}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product; hence $\sum_{k=1}^{n}x^{k}=x\,\frac{1-x^{n}}{1-x}$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive. Now $x^{n}\to0$ by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences §geometric, and the constant sequence $(1)_{n\in\mathbb{N}}$ converges to $1$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant, so by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic first $1-x^{n}\to1$, then $\frac{1}{1-x}(1-x^{n})\to\frac{1}{1-x}$ and $\frac{x}{1-x}(1-x^{n})\to\frac{x}{1-x}$. Hence $\sum_{k=1}^{\infty}x^{k-1}$ and $\sum_{k=1}^{\infty}x^{k}$ converge, with sums $\frac{1}{1-x}$ and $\frac{x}{1-x}$. Now let $|r|<1$. Taking $x=r$ gives $1-r\neq0$ and the convergence of both series with the stated sums. As $\big||r|\big|=|r|<1$, taking $x=|r|$ shows that $\sum_{k=1}^{\infty}|r|^{k-1}$ and $\sum_{k=1}^{\infty}|r|^{k}$ converge; since $|r^{k-1}|=|r|^{k-1}$ and $|r^{k}|=|r|^{k}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign, these are the series $\sum_{k=1}^{\infty}|r^{k-1}|$ and $\sum_{k=1}^{\infty}|r^{k}|$, so both series converge absolutely.

Clause geometric-tail. Let $|r|<1$ and $n\in\mathbb{N}$. By the paragraph on clause geometric, taken with $x=r$, $1-r\neq0$, $\sum_{k=1}^{n}r^{k}=r\,\frac{1-r^{n}}{1-r}$ and $\sum_{k=1}^{\infty}r^{k}=\frac{r}{1-r}$. As $r\,r^{n}=r^{1}r^{n}=r^{1+n}=r^{n+1}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents, we get $r(1-r^{n})=r-r^{n+1}$, so $\sum_{k=1}^{n}r^{k}=\frac{r-r^{n+1}}{1-r}$. Subtracting, $\sum_{k=1}^{\infty}r^{k}-\sum_{k=1}^{n}r^{k}=\frac{r}{1-r}-\frac{r-r^{n+1}}{1-r}=\frac{r^{n+1}}{1-r}$.