Proof of Cesaro Means of a Convergent Real Sequence Converge to Its Limit

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

By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, $\mathbb{R}$ is an ordered field; the rules of arithmetic, order and absolute value used without further comment are those of Rules of Arithmetic and Order in an Ordered Field, which the setting carries, together with the field laws of Commutative Rings §ring. Natural numbers are read in $\mathbb{R}$ as in The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §identification; this reading is that of The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §embedding, and sums, products and the order agree whether formed in $\mathbb{N}$ or in $\mathbb{R}$, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement. For every $n\in\mathbb{N}$ we have $0<n$ in $\mathbb{R}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals; hence $n\neq0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization, so $n^{-1}$ and $\frac1n$ are defined by Negatives, Differences, Reciprocals and Quotients §reciprocal, and $\frac1n=1\cdot n^{-1}=n^{-1}$ by that clause and the commutativity and unit law of Commutative Rings §ring. Thus $n\cdot\frac1n=1$ by Negatives, Differences, Reciprocals and Quotients §reciprocal, $\frac1n\cdot n=1$ by commutativity, and $0<\frac1n$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. For $q\in\mathbb{N}$ and a map $c$ from a set containing $[q]$ to $\mathbb{R}$, the sum $\sum_{k=1}^{q}c_{k}$ is, through the setting, the sum of Sums and Products over a Finite Set and over an Interval §sums and Sums and Products over a Finite Set and over an Interval §intervals, that is, by Sums and Products over a Finite Set and over an Interval §subsets, the sum of the restriction $c|_{[q]}$; every sum below is of this kind, and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality is a lemma about these sums. By Sums and Products over a Finite Set and over an Interval §intervals they agree with the iterated sums of Iterated Operations: Finite Sums and Finite Products §iterated, taken of the restriction to $[n]$ as in Iterated Operations: Finite Sums and Finite Products §restriction, so Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting applies to them. Put $b_{k}=|a_{k}-a|$ for $k\in\mathbb{N}$, so $0\le b_{k}$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, and $s_{n}=\frac1n\sum_{k=1}^{n}a_{k}$ for $n\in\mathbb{N}$; the sequence $(s_{n})$ of the clause cesaro is the map $n\mapsto s_{n}$ on $\mathbb{N}$, unique by Sets and Maps: Ordinary Notation §maps.

Step 1 (Comparison with the averaged errors). Let $n\in\mathbb{N}$. By Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §constant, $\sum_{k=1}^{n}a=na$, so $a=\frac1n\sum_{k=1}^{n}a$, as $\frac1n\cdot(n\cdot a)=(\frac1n\cdot n)\cdot a=a$ by associativity, $\frac1n\cdot n=1$ (shown above), commutativity and the unit law. Hence, by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §difference and distributivity,

$$s_{n}-a=\frac1n\Bigl(\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}a\Bigr)=\frac1n\sum_{k=1}^{n}(a_{k}-a),\qquad\text{so}\qquad|s_{n}-a|\le\frac1n\sum_{k=1}^{n}b_{k}\tag{$*$}$$

by $|xy|=|x||y|$ (Rules of Arithmetic and Order in an Ordered Field §absolute-value), $\bigl|\frac1n\bigr|=\frac1n$ (Absolute Value in an Ordered Field §absolute-value, as $0<\frac1n$), Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §triangle and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling.

Step 2 (Order of choice). Let $\varepsilon\in\mathbb{R}$ be positive; then $\varepsilon/2$ is positive by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. First, since $a_{n}\to a$, by Convergent Sequences of Real Numbers §converges take $K\in\mathbb{N}$ with $b_{k}<\varepsilon/2$ for every $k\in\mathbb{N}$ with $k\ge K$, and put $A=\sum_{k=1}^{K}b_{k}$, which is nonnegative by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §nonnegative. Next, $\varepsilon\neq0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization, so $\varepsilon^{-1}$ is defined by Negatives, Differences, Reciprocals and Quotients §reciprocal; by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean, take $M\in\mathbb{N}$ with $(2A)\cdot\varepsilon^{-1}<M$. As $0<\varepsilon$, Rules of Arithmetic and Order in an Ordered Field §order-product gives $((2A)\cdot\varepsilon^{-1})\cdot\varepsilon<M\cdot\varepsilon$, and $((2A)\cdot\varepsilon^{-1})\cdot\varepsilon=(2A)\cdot(\varepsilon\cdot\varepsilon^{-1})=(2A)\cdot1=2A$ by the associativity, commutativity and unit law of Commutative Rings §ring and by Negatives, Differences, Reciprocals and Quotients §reciprocal; so $2A<M\cdot\varepsilon$. Put $N=M+K$. By Arithmetic and Order of the Natural Numbers §difference and Arithmetic and Order of the Natural Numbers §commutative, $K<N$ and $M<N$.

Step 3 (Tail bound). Let $n\in\mathbb{N}$ with $n\ge N$. Then $K<n$, since $K<N$ and $N\le n$: by Arithmetic and Order of the Natural Numbers §partial-order, $N\le n$ means $N<n$ or $N=n$, and in the first case $K<n$ by the strict transitivity in the same clause. So by Arithmetic and Order of the Natural Numbers §difference there is $p\in\mathbb{N}$ with $n=K+p$; then $n=p+K$ by Arithmetic and Order of the Natural Numbers §commutative, so $p<n$ by Arithmetic and Order of the Natural Numbers §difference again. Addition in $\mathbb{R}$ is associative by Commutative Rings §ring, so by Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting

$$\sum_{k=1}^{n}b_{k}=A+\sum_{i=1}^{p}b_{K+i}.$$

For each $i\in[p]$ we have $K<K+i$ by Arithmetic and Order of the Natural Numbers §difference, so $K\le K+i$ and $b_{K+i}<\varepsilon/2$; hence $\sum_{i=1}^{p}b_{K+i}\le\sum_{i=1}^{p}\frac{\varepsilon}{2}=p\cdot\frac{\varepsilon}{2}\le n\cdot\frac{\varepsilon}{2}$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison, Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §constant and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling, as $p\le n$ also in $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement. Thus $\sum_{k=1}^{n}b_{k}\le A+n\cdot\frac{\varepsilon}{2}$ by Rules of Arithmetic and Order in an Ordered Field §order-sum.

Step 4 (Conclusion). Let $n\in\mathbb{N}$ with $n\ge N$. First, $M<n$, from $M<N$ and $N\le n$ by Arithmetic and Order of the Natural Numbers §partial-order as in Step 3, so $M\le n$ by the same clause, and also in $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement. As $0<\varepsilon$, Rules of Arithmetic and Order in an Ordered Field §order-product gives $M\cdot\varepsilon\le n\cdot\varepsilon$, so $2A<n\cdot\varepsilon$ by Step 2 and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. As $0<n^{-1}$ (shown at the start), Rules of Arithmetic and Order in an Ordered Field §order-product gives $(2A)\cdot n^{-1}<(n\cdot\varepsilon)\cdot n^{-1}$; here $(n\cdot\varepsilon)\cdot n^{-1}=\varepsilon\cdot(n\cdot n^{-1})=\varepsilon\cdot1=\varepsilon$ and $(2A)\cdot n^{-1}=2\cdot(A\cdot n^{-1})=2\cdot\frac{A}{n}$, by the associativity, commutativity and unit law of Commutative Rings §ring and by Negatives, Differences, Reciprocals and Quotients §reciprocal, which gives $n\cdot n^{-1}=1$ and $\frac{A}{n}=A\cdot n^{-1}$. So $2\cdot\frac{A}{n}<\varepsilon$. Next, $0<2$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving, so $2\neq0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization, $2^{-1}$ is defined by Negatives, Differences, Reciprocals and Quotients §reciprocal, and $0<2^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. By Rules of Arithmetic and Order in an Ordered Field §order-product again, $\bigl(2\cdot\frac{A}{n}\bigr)\cdot2^{-1}<\varepsilon\cdot2^{-1}=\frac{\varepsilon}{2}$, and $\bigl(2\cdot\frac{A}{n}\bigr)\cdot2^{-1}=\frac{A}{n}\cdot(2\cdot2^{-1})=\frac{A}{n}\cdot1=\frac{A}{n}$ by the same laws; so $\frac{A}{n}<\frac{\varepsilon}{2}$. Now, by ($*$), Step 3 and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling with $0\le\frac1n$ (as $0<\frac1n$) and commutativity,

$$|s_{n}-a|\le\frac1n\Bigl(A+n\cdot\frac{\varepsilon}{2}\Bigr)=\frac{A}{n}+\frac1n\cdot\Bigl(n\cdot\frac{\varepsilon}{2}\Bigr)=\frac{A}{n}+\frac{\varepsilon}{2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

Here the first equality is distributivity together with $\frac1n\cdot A=A\cdot n^{-1}=\frac{A}{n}$, as $\frac1n=n^{-1}$ (shown at the start); the second is $\frac1n\cdot\bigl(n\cdot\frac{\varepsilon}{2}\bigr)=\bigl(\frac1n\cdot n\bigr)\cdot\frac{\varepsilon}{2}=1\cdot\frac{\varepsilon}{2}=\frac{\varepsilon}{2}$, by associativity, $\frac1n\cdot n=1$ (shown at the start), commutativity and the unit law; the strict step is Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum, applied to $\frac{A}{n}<\frac{\varepsilon}{2}$ and $\frac{\varepsilon}{2}\le\frac{\varepsilon}{2}$, combined with the preceding inequality by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed; and the last equality is Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. So for every positive $\varepsilon\in\mathbb{R}$ there is $N\in\mathbb{N}$ with $|s_{n}-a|<\varepsilon$ for every $n\in\mathbb{N}$ with $n\ge N$; that is, $s_{n}\to a$ by Convergent Sequences of Real Numbers §converges, which is the clause cesaro.