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

Each result cited is universally quantified over the data in its own statement. Elementary arithmetic and order facts in $\mathbb{R}$, the properties of $|\cdot|$, the properties of finite sums in Properties of Finite Sums and the Archimedean property are those put in force by The Real Numbers: Standing Notation and Background §background; natural numbers are read in $\mathbb{R}$ as in The Real Numbers: Standing Notation and Background §numbers. Put $b_{k}=|a_{k}-a|$ and $s_{n}=\tfrac1n\sum_{k=1}^{n}a_{k}$.

Step 1 (Two facts about finite sums, and a consequence). For every $n\in\mathbb{N}$: (i) $\sum_{k=1}^{n}a=na$; (ii) $\bigl|\sum_{k=1}^{n}(a_{k}-a)\bigr|\le\sum_{k=1}^{n}b_{k}$. Both hold for $n=1$ by the first formula of claim 1 of Properties of Finite Sums, and pass from $n$ to $n+1$ by its recursion formula: $\sum_{k=1}^{n+1}a=na+a=(n+1)a$, and $\bigl|\sum_{k=1}^{n+1}(a_{k}-a)\bigr|\le\bigl|\sum_{k=1}^{n}(a_{k}-a)\bigr|+|a_{n+1}-a|\le\sum_{k=1}^{n+1}b_{k}$ by the triangle inequality, claim 5 of Properties of the Absolute Value in an Ordered Field. So both hold for every $n$ by Principle of Induction for the Natural Numbers, applied to the set of $n\in\mathbb{N}$ for which (i) and (ii) hold. By claims 2 and 3 of Properties of Finite Sums and (i),

by (ii), as $\tfrac1n$ is positive.

Step 2 (Choice of indices). Let $\varepsilon$ be positive. First, by Limit of a Sequence of Real Numbers, choose $K\in\mathbb{N}$ with $b_{k}<\varepsilon/2$ for every $k\ge K$, and put $A=\sum_{k=1}^{K}b_{k}$, which is nonnegative by claim 5 of Properties of Finite Sums. Then, by claim 1 of The Archimedean Property of the Real Numbers, choose $N_{0}\in\mathbb{N}$ with $2A/\varepsilon<N_{0}$, and put $N=N_{0}+K$, so that $K<N$ and $2A/\varepsilon<N$.

Step 3 (Tail bound). For every $n\ge K$,

Indeed, the set of natural numbers $j$ for which this holds with $n=K+j-1$ contains $1$, since $\sum_{k=1}^{K}b_{k}=A\le A+K\varepsilon/2$; and if it contains $j$, then with $n=K+j-1$ the recursion of claim 1 of Properties of Finite Sums and $b_{n+1}<\varepsilon/2$ (as $n+1\ge K$) give $\sum_{k=1}^{n+1}b_{k}\le A+n\varepsilon/2+\varepsilon/2=A+(n+1)\varepsilon/2$, so it contains $j+1$. By Principle of Induction for the Natural Numbers it is all of $\mathbb{N}$, and every $n\ge K$ has the form $K+j-1$ with $j\in\mathbb{N}$.

Step 4 (Conclusion). Let $n\ge N$. Then $n\ge K$, and by ($*$) and Step 3,

using $0\le A$, $0<N\le n$, and $A/N<\varepsilon/2$, which is $2A/\varepsilon<N$ multiplied by the positive number $\varepsilon/(2N)$. By Limit of a Sequence of Real Numbers, $(s_{n})$ converges to $a$.