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Construct partial sums $s_n$ from alternating blocks of zero and one, with lengths $1,3,9,27,\dots$. Block $j$, beginning with $j=0$, has length $3^j$ and value zero when $j$ is even, one when $j$ is odd. Define a series by $a_0=s_0$ and $a_n=s_n-s_{n-1}$ for $n\ge 1$; the telescoping series identity recovers these chosen partial sums. They remain in $[0,1]$.

At the end of odd block $2m+1$, the Cesàro mean is exactly $3/4$. At the end of even block $2m$, it isThese two subsequences have different subsequential limits, so the series has no first-order Cesàro sum. A later block remains a substantial fraction of everything averaged so far. Boundedness alone therefore does not ensure a stable average.

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

The Cesàro mean of the first $n\ge 1$ terms of a sequence is their arithmetic mean. It is a cumulative average: increasing $n$ includes the whole initial segment, rather than a fixed-length sliding window. For $a_k=(-1{)}^k$, the even-length averages are zero and the odd-length averages are $-1/n$, so they tend to zero even though the sequence does not converge.

For a periodic sequence with period $p\ge 1$ and one-cycle arithmetic mean $M$, split an initial segment into full cycles and a remainder of fewer than $p$ terms. The full cycles contribute exactly $M$ per term; the remainder contributes a fixed bounded error divided by $n$. Thus the Cesàro means tend to $M$. The cycle $0,2,1$ in the bounded partial sums need not be Cesàro summable illustration has arithmetic mean one. Treating it as a sequence of partial sums gives a series with first-order Cesàro sum one; averaging a sequence and summing its terms remain different operations.

Cesàro summation assigns a value to a series when the arithmetic means of its partial sums approach a finite limit of a sequence. Here we use the first-order method, often written $(C,1)$. For a real or complex series beginning at index zero, defineIt is Cesàro summable to $L$ when ${\sigma }_N\to L$. The averaging is applied to the partial sums $s_n$, rather than to the original terms $a_n$. This distinction is essential: the terms can average to zero while their accumulated sum keeps growing. ^[ref]

For Grandi's series $1-1+1-1+\cdots$, the partial sums alternate between one and zero. The series is not a convergent series, but the two values occupy asymptotically equal proportions of the averaging window. In the declared index convention,so its first-order Cesàro sum is $1/2$. This names a particular summation method; it does not turn the oscillating partial sums into an ordinarily convergent sequence.

Why is this useful? An oscillating approximation can retain stable information when its cumulative average settles. Because the Cesàro theorem for convergent sequences preserves an existing limit of a sequence, this adds some new cases without changing the sum of any ordinarily convergent series. For Fourier series, averaging the Fourier partial sums gives a Fejér sum; the existing Fejér kernel and approximation articles explain its effect on reconstruction. Averaging does not guarantee success for every oscillating sequence, as the bounded partial sums need not be Cesàro summable example shows.

One historical motivation was the multiplication of series. In 1918, A. F. Andersen recalled Ernesto Cesàro's product theorem and cited his 1890 publication at the start of a study of products involving Cesàro summable series. This connects the theory with preserving useful operations on sums, as well as with assigning a value to an isolated example. His paper treats several orders of summability; the examples here use only $(C,1)$.^[ref]

For a formal cosine series ${\sum }_j{a}_j\cos (j!t)$ with uniformly bounded coefficients, the Fourier Cesàro mean at $N_J$ differs from its ordinary partial sum by $-(N_J+1{)}^{-1}{\sum }_{j\le J}j!a_j\cos (j!t)$. Since ${\sum }_{j\le J}j!\le 2J!$, the difference is $O(1/J)$ uniformly in $t$. Thus any divergent behavior of these ordinary cutoff sums persists along the same Cesàro subsequence. Averaging by the sparse coefficient index is a different summation operation from averaging the actual Fourier frequency cutoffs.

If a real or complex sequence converges to $a$, its Cesàro means have the same limit of a sequence. This is why Cesàro summation is a regular summation method when applied to the partial sums of a series.

Proof outline. Given $\epsilon >0$, choose $K$ so that $|a_k-a|<\epsilon$ for $k>K$. The triangle inequality gives, for $n>K$,The finite initial contribution tends to zero, and the remaining contribution is at most $\epsilon$. The converse fails: the alternating sequence $(-1{)}^n$ has Cesàro means tending to zero while its terms do not converge.

For a strictly stationary process on a finite alphabet, put $a_1=H(X_1)$ and $a_n=H(X_n\mid X_1,\dots ,X_{n-1})$ for $n\ge 2$. Each $a_n$ is the average remaining uncertainty in the next symbol after its history is known. More history can improve prediction, so this sequence is nonincreasing:The inequality uses conditioning reduces entropy; the last equality uses strict stationarity. Nonnegative $a_n$ therefore have a limit. By the chain rule for information entropy,The Cesàro mean converges to the same limit, proving the equality with the entropy rate. This is a statement about average information entropy. A constant almost-sure limit of the realized information content requires the further hypotheses in the Shannon-McMillan-Breiman theorem; stationarity without ergodicity in an information source explains the distinction.

The Cesàro means of the Fourier series of a continuous periodic function converge uniformly to that function. The Fejér kernel is nonnegative with total mass one after normalization, and its mass outside every fixed neighborhood of zero tends to zero. Uniform continuity controls the near contribution, while the vanishing tail controls the rest. For an $L^1$ function the same positive-kernel argument gives an $L^1$ contraction, which is useful even when pointwise behavior is unavailable.

For independent random variables with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$, writeThen $\Vert Y_N-m_N{\Vert }_2^2=N^{-2}{\sum }_{k\le N}\mathrm{Var}(X_k)=O(N^{-1})$. Hence $Y_N$ converges in $L^2$ exactly when the scalar Cesàro means $m_N$ converge.

Let $C{a}_n=n^{-1}{\sum }_{j=1}^n{a}_j$ be the Cesàro mean of a sequence. If $|a_n|\le Cn$ for every positive integer $n$, the triangle inequality gives $|C{a}_n|\le C(n+1)/2\le Cn$. Mathematical induction repeats this estimate with the same constant for every iterate. An eventual $O(n)$ bound can be made global by increasing $C$ to absorb the finitely many initial terms. Consequently iterated Cesàro means preserve linear growth in Big O notation; this bound alone does not assert convergence or summability.

With $\hat{f}(k)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(s)e^{-iks}ds$, the classical Fourier partial sum is $s_N^f(t)={\sum }_{|k|\le N}\hat{f}(k)e^{ikt}$. Its Cesàro mean isFor the uniform Fejér theorem, extend the continuous endpoint-matching $f$ periodically. The given Fejér kernel is nonnegative, even and has total integral $\pi$. ThusGiven $\epsilon >0$, uniform continuity supplies $\delta >0$ so the bracket has modulus less than $\epsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\epsilon$. On the remainder, $F_N(u)\le [2(N+1){\sin }^2(\delta /2){]}^{-1}$, and the bracket is at most $2\Vert f{\Vert }_{\infty }$. HenceLetting $\epsilon$ decrease to zero proves uniform convergence of the Fejér sums.

The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt{j}}$ and consider its formal trigonometric series. For $t=\pi p/q$, there is $J_0$ such that $\cos (j!t)=1$ for every $j\ge J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes toThe exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m{)}^2$ and $((2m+1)\pi {)}^2$. Thus the ordinary sums diverge at every $t\in \pi \mathbb{Q}$.

But their genuine Fourier Cesàro means also diverge there. Indeed,Since $|a_j|\le 2$ and ${\sum }_{j=0}^{J}j!\le 2J!$ for $J\ge 3$, its absolute value is at most $4/(J+1)\to 0$, uniformly in $t$. Thereforewhich has the same incompatible limit points. This is an instance of Cesàro sums at factorial cutoffs track ordinary sums. The printed Cesàro-convergence claim is false. Averaging the sparse series by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.

There is a further issue: no $L^1$ function has exactly this Fourier series. Here is a direct proof using strongly lacunary polynomial L1-L2 bound. For a trigonometric polynomial $p$ supported on $\{\pm j!:j\ge 3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi )$ and bounding the three possible pairings givesThe Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$, hence $\Vert p{\Vert }_2\le \sqrt{3}\Vert p{\Vert }_1$.

If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\Vert {\sigma }_N^f{\Vert }_1\le \Vert f{\Vert }_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a polynomial of the type above. Its $L^2$ norms would therefore be uniformly bounded. Orthogonality then forces ${\sum }_{j\ge 3}|a_j{|}^2<\infty$, by fixing any finite coefficient block and letting $N\to \infty$. However,so that sum diverges. This contradiction disproves the stated existence premise.

Even considered only formally, these coefficients rule out an $L^2$ function by Bessel's inequality, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted series either. The asserted $L^1$ function does not exist.

Use the dominated form of the real Hahn-Banach theorem. Let $p:V\to \mathbb{R}$ be a sublinear functional on a real vector space, meaning $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for $t\ge 0$. If a linear functional $f$ on a vector subspace $M$ satisfies $f\le p$ there, then it has a linear extension $F$ to $V$ with $F\le p$ everywhere.

Here is the one-dimensional extension step. For $z\notin M$, the value $a=F(z)$ must satisfyThis interval is nonempty: for any $m,n\in M$,so every candidate lower bound is at most every candidate upper bound. The candidates with $m=n=0$ also bound the endpoints by finite numbers, so a real $a$ can be chosen. Define $F(m+tz)=f(m)+ta$. For $t>0$, the upper bound, applied to $m/t$, gives $F(m+tz)\le p(m+tz)$ after multiplication by $t$. For $t<0$, the lower bound applied to $m/(-t)$ gives the same conclusion after multiplication by $-t$; $t=0$ is the original hypothesis. Thus the extension remains dominated.

Order all dominated extensions by extension of their domains and values. Every chain has the dominated linear union as an upper bound. The Zorn lemma supplies a maximal extension. If its domain were not all of $V$, the step just proved would enlarge it, a contradiction. This proves the real Hahn-Banach extension theorem. In particular, taking $p(x)=\Vert f\Vert \Vert x\Vert$ on a normed vector space and applying the domination to both $x$ and $-x$ yields the usual norm-preserving extension of a bounded linear functional.

For generalized limits, take $V={\ell }^{\infty }(\mathbb{N};\mathbb{R})$, the l-infinity sequence space, and $M=c$, the convergent sequence space. Let $f(x)={\lim }_n{x}_n$ on $c$ and $p(x)={\mathrm{limsup}}_n{x}_n$ on $V$. This $p$ is finite and sublinear, and agrees with $f$ on $c$. The dominated extension $L$ therefore satisfiesThus a generalized limit exists, withIt is positive, agrees with the ordinary limit on every convergent sequence, and satisfies $|L(x)|\le \Vert x{\Vert }_{\infty }$. Since $L(1)=1$, its operator norm is exactly one.

If generalized limits are additionally required to be shift invariant, the same theorem gives the stronger Banach limit. Replace the dominating functional byIt is again sublinear and agrees with $f$ on $c$, by the Cesàro theorem for convergent sequences. The resulting $L_C$ lies between the lower and upper limits of the Cesàro means. These in turn lie between the ordinary lower and upper limits of a bounded sequence, so $L_C$ is a positive, norm-one extension of the ordinary limit. For the left shift on bounded sequences $S$, the means of $Sx-x$ equal $(x_{N+1}-x_1)/N\to 0$. Applying domination to both signs forces $L_C(Sx-x)=0$. HenceThis Cesàro construction of a Banach limit proves existence even under the stronger convention, rather than leaving shift invariance implicit.

The series ${\sum }_{n\ge 1}{a}_n$ converges to $A$ when its partial sums $s_n$ have limit of a sequence $A$: for every $\epsilon >0$, there is $N$ such that $|s_n-A|<\epsilon$ for all $n\ge N$.

Its averages are the Cesàro means of the sequence of partial sums. To prove their convergence, choose $N$ with $|s_k-A|<\epsilon /2$ for $k\ge N$. ThenThe finite first sum divided by $n$ tends to zero. This proves the Cesàro theorem for convergent sequences in this setting, andThe converse fails. Take $a_n=(-1{)}^{n+1}$, so $s_n=1$ for odd $n$ and $s_n=0$ for even $n$. The partial sums do not converge, butThus the alternating series is Cesàro summable without having an ordinary sum.

For a weakly stationary process with mean $m$, the autocovariance function is ${\gamma }_k=\mathbb{E}[(X_{t+k}-m)(X_t-m)]$, independent of $t$. The positive definiteness of these covariances gives a unique finite positive spectral measure of a stationary time series on the frequency circle. Its cumulative function is the time-series spectral distribution $F$, with total mass ${\gamma }_0$, andUse one representative of the identified endpoints $-\pi$ and $\pi$ when specifying atoms. For a real-valued process the measure is symmetric, so the imaginary part integrates to zero. If it has a time-series spectral density, thenThe second identity is pointwise for an absolutely summable covariance sequence, as will apply to part (a). Existence of a density alone does not guarantee ordinary pointwise convergence of its Fourier series; in general the reconstruction can be interpreted by its Cesàro means in $L^1$. This fixes the frequency normalization for all three calculations below.

The Baire category theorem states that, in a nonempty complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty open set cannot be covered by countably many nowhere dense sets.

For the Uniform boundedness principle, let $X$ be a Banach space, $Y$ a normed vector space, and $(T_{\alpha })$ a family of bounded linear operators for which ${\sup }_{\alpha }\Vert T_{\alpha }x\Vert <\infty$ for every $x\in X$. PutEach $F_n$ is closed, being an intersection of inverse images of closed balls, and $X={\bigcup }_n{F}_n$. By the Baire category theorem, some $F_n$ has interior. Choose a closed ball $\overline{B}(x_0,r)\subseteq F_n$ with $r>0$. For $\Vert h\Vert \le r$, both $x_0$ and $x_0+h$ belong to $F_n$, and linearity gives $\Vert T_{\alpha }h\Vert \le 2n$. Scaling yieldsCompleteness is needed for the domain, not the codomain.

We next prove the Silverman-Toeplitz theorem. Write $c$ for the real convergent sequence space with the supremum norm. It is a Banach space: the l-infinity sequence space is complete by coordinatewise passage to limits of norm-Cauchy sequences, and $c$ is closed in it. Indeed, if $x^{(m)}\to x$ uniformly and each $x^{(m)}$ converges, approximate $x$ within $\epsilon /3$ by one $x^{(m)}$ and make that sequence's tail oscillation smaller than $\epsilon /3$. Then the tail of $x$ is Cauchy and therefore converges in $\mathbb{R}$.

Suppose $A$ is a regular summability matrix. Fix a row $i$ and consider the finite bounded linear functionalsFor every $x\in c$ their values converge, by the definition of regularity, and are therefore bounded. Their operator norms are exactly ${\sum }_{j\le N}|a_{ij}|$: the upper bound is the triangle inequality, and the finite sign sequence $x_j=\mathrm{sgn}(a_{ij})$ for $j\le N$, zero afterwards, attains it. Applying the Uniform boundedness principle to this row shows ${\sum }_j|a_{ij}|<\infty$.

Consequently the full row functional $A_i:c\to \mathbb{R}$ is continuous and has norm ${\sum }_j|a_{ij}|$, using the same finite sign tests and then letting $N\to \infty$. Regularity makes $(A_i(x){)}_i$ convergent, hence bounded, for every $x\in c$. A second application of the Uniform boundedness principle gives a common row bound $M$. Testing regularity on a coordinate vector $e_j$ and on the constant-one sequence gives, respectively, convergence to zero down each column and convergence of row sums to one. Thus the necessary conditions are

Conversely assume these conditions and let $x_j\to L$. The sequence is bounded, so every row sum defining $A_i(x)$ converges absolutely. WriteGiven $\epsilon >0$, choose $N$ such that $|x_j-L|<\epsilon /(M+1)$ for $j>N$. The tail of the first sum has modulus at most $M\epsilon /(M+1)<\epsilon$, uniformly in $i$. Its finite head tends to zero by columnwise convergence, and $L(s_i-1)\to 0$. Since $\epsilon$ is arbitrary, $A_i(x)\to L$. The three conditions are necessary and sufficient for regularity.

For the averaging matrix $B$, every absolute row sum and every row sum equals $1$. A fixed column eventually has entry $1/r\to 0$. Thus $B$ is a regular summability matrix, and its transformed values are the Cesàro means. To exhibit a bounded sequence without a Cesaro limit, put $N_0=0$, $N_m=2^{2^m}$ for $m\ge 1$, and setAt the end of the $m$th block,The averages at even block endpoints approach $1$, while those at odd endpoints approach $-1$. Hence $w$ is bounded and has no $B$-limit.

For an arbitrary regular summability matrix $A$, use a gliding hump argument; this also handles negative entries. Put $N_0=0$ and $i_0=0$. Inductively choose $i_m>i_{m-1}$ so large thatThis is possible by the finite-column convergence and convergence of the row sums to one. Since the selected row is absolutely summable, choose $N_m>N_{m-1}$ withDefine a single bounded sequence on the successive disjoint blocks byusing $\mathrm{sgn}(0)=0$. The row's absolute mass on its own block is greater than $3/4-1/8=5/8$, while all coordinates outside that block contribute a term of modulus less than $1/8$, irrespective of the signs subsequently assigned. ThereforeEven and odd selected rows cannot tend to the same limit. Every row series does converge absolutely because $\Vert y{\Vert }_{\infty }\le 1$. This proves bounded divergence for every regular summability matrix.

The row bound makes $T_A:{\ell }^{\infty }\to {\ell }^{\infty }$, $T_{A}x=(A_i(x){)}_i$, a bounded linear operator with $\Vert T_A\Vert \le M$. The summability domain of a regular matrix isSince $c$ is closed, this is a closed linear subspace. It is proper by the constructed $y$. A proper vector subspace of a normed vector space has empty interior: a ball in the subspace can be translated to a ball about zero, and scaling that ball would put every vector in the subspace. ThusFor a countable family of regular matrices, the Baire category theorem on the Banach space ${\ell }^{\infty }$ givesAny vector in this complement has none of the indicated matrix limits. The matrices in this clause must be regular, as in the preceding discussion; without that implicit qualification even a single zero matrix sums every bounded sequence to zero.

Finally take any bounded sequence $x$. By the Bolzano-Weierstrass theorem it has a subsequence $x_{n_i}\to L$ with $n_i$ strictly increasing. Define the selector matrixIts absolute row sums and row sums are $1$, and each fixed column is eventually zero. The Silverman-Toeplitz theorem makes it regular, while $A_i(x)=x_{n_i}\to L$. Thus every bounded sequence is summable by some regular matrix depending on that sequence. This is subsequence selection as regular matrix summability; it reverses the quantifier order of the impossibility result for a fixed matrix.

A proper filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes $\varnothing$, is closed under finite intersections, and is upward closed under inclusion. In particular $X\in \mathcal{F}$. In a topological space, convergence of a filter $\mathcal{F}\to x$ means that every neighbourhood of $x$ belongs to $\mathcal{F}$. An ultrafilter is a proper filter maximal under inclusion; equivalently, it contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To justify the equivalence, if $A\notin \mathcal{U}$ and every member of $\mathcal{U}$ met $A$, adjoining $A$ would generate a larger proper filter. Maximality therefore supplies $F\in \mathcal{U}$ with $F\cap A=\varnothing$, and upward closure gives $X\setminus A\in \mathcal{U}$. Both complementary sets cannot belong to a proper filter. Conversely, a filter with this dichotomy cannot acquire any new set without also containing its complement, which would give an empty intersection.

The ultrafilter characterization of compact Hausdorff spaces, stated without proof, is: a topological space is compact and Hausdorff exactly when every ultrafilter on it converges to a unique point.

For a finite set partition $X=A_1\cup \cdots \cup A_n$, if no cell belonged to $\mathcal{U}$, the ultrafilter dichotomy would put all $X\setminus A_j$ in $\mathcal{U}$. Their finite intersection is empty, a contradiction. Thus some $A_j\in \mathcal{U}$, and at most one cell can belong because distinct cells are disjoint. The converse is: a proper filter on a set meeting every finite partition by containing one cell is an ultrafilter. Apply the partition hypothesis to $X=A\cup (X\setminus A)$ to obtain the dichotomy, then the maximality argument above. Empty cells can be omitted, and the cases $A=\varnothing$ or $A=X$ are automatic.

For a function $f:X\to Y$, the image of a filter isSince preimages preserve inclusions and finite intersections, this is a proper filter on a set $Y$. Equivalently, it is the filter generated by $f(F)$ for $F\in \mathcal{F}$: if $f^{-1}(B)\in \mathcal{F}$, then its image is contained in $B$, while $f(F)\subseteq B$ implies $F\subseteq f^{-1}(B)$. For an ultrafilter $\mathcal{U}$, the identity $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ shows that $f_{\ast }\mathcal{U}$ contains exactly one of $B,Y\setminus B$. Therefore the pushforward filter of an ultrafilter is an ultrafilter.

Now let $x\in {\ell }^{\infty }$, the l-infinity sequence space, and put $a_x(n)=n^{-1}{\sum }_{j=1}^n{x}_j$. All these Cesàro means lie in the closed disk $K_x=\{z:|z|\le \Vert x{\Vert }_{\infty }\}$, a compact Hausdorff space. The pushforward filter $a_{x\ast }\mathcal{U}$ contains $K_x$; restricting it to that disk and applying the ultrafilter theorem gives a unique limitFor real sequences the disk can be replaced by a closed interval. This is the ultrafilter construction of a Banach limit.

The identities $a_{\alpha x+\beta y}(n)=\alpha a_x(n)+\beta a_y(n)$ give linearity of $l$. Explicitly, the two sets on which $|a_x(n)-l(x)|<\delta$ and $|a_y(n)-l(y)|<\delta$ both belong to $\mathcal{U}$, hence so does their intersection. On that intersection the difference from $\alpha l(x)+\beta l(y)$ is at most $(|\alpha |+|\beta |)\delta$, proving $l(\alpha x+\beta y)=\alpha l(x)+\beta l(y)$. If $x_n\ge 0$, its means lie in $[0,\Vert x{\Vert }_{\infty }]$, so $l(x)\ge 0$. Thus $l$ is a positive linear functional. The bound $|l(x)|\le \Vert x{\Vert }_{\infty }$ gives $\Vert l\Vert \le 1$, while the constant-one sequence satisfies $l(1)=1$. Hence $\boxed{\Vert l\Vert =1}$.

Because $\mathcal{U}$ contains no finite sets, its dichotomy puts every cofinite set in $\mathcal{U}$. If $x_n\to L$ ordinarily, the Cesàro theorem for convergent sequences follows directly by splitting the sum: for a fixed $N$ with $|x_j-L|<\epsilon$ for $j>N$,The finite head tends to zero, so $a_x(n)\to L$ ordinarily. Every sufficiently long tail belongs to $\mathcal{U}$, making its ultralimit also $L$. Thus $\boxed{x_n\to L⟹l(x)=L}$.

Finally, for the left shift on bounded sequences $S$,Taking ultralimits along the same cofinite-containing ultrafilter gives $\boxed{l(Sx)=l(x)}$. This establishes every defining property of a Banach limit.

Let $d_j=a_{j+1}-a_j$ and suppose $d_j\to L$. Telescoping givesTo justify the Cesàro mean limit directly, choose $J$ so that $|d_j-L|<\epsilon$ for $j\ge J$. The finite initial sum contributes at most $n^{-1}{\sum }_{j<J}|d_j-L|$, which tends to zero. The remaining terms contribute at most $\epsilon$, and replacing $(n-1)L/n$ by $L$ contributes $|L|/n$. Consequently $\boxed{a_n/n\to L}$.

Because $(a_n)$ is a monotone sequence, every one of its first $n$ terms is at most $a_n$, soOn the other hand, every term from $a_n$ through $a_{2n}$ is at least $a_n$, and henceBoth the lower bound and upper bound tend to $x$ because the Cesàro means converge to $x$. The squeeze theorem gives

Suppose $a_n\to L$. Given $\epsilon >0$, choose $N$ such that $|a_k-L|<\epsilon /2$ for $k>N$. ThenThe first term tends to zero because its numerator is fixed, and the second is at most $\epsilon /2$. Thus the Cesàro mean tends to $L$.

The converse is false. For $a_n=(-1{)}^n$, the Cesaro means tend to $0$, while the original sequence alternates between $-1$ and $1$ and does not converge.

The strong law of large numbers states that for independent identically distributed integrable random variables,To prove it, set $Y_n=X_n{1}_{\{|X_n|\le n\}}$. The tail-sum formula gives ${\sum }_n\mathbb{P}(X_n\ne Y_n)<\infty$, so the Borel-Cantelli lemmas make the two sequences eventually equal. AlsoThe Kolmogorov convergence theorem implies that ${\sum }_n(Y_n-\mathbb{E}{Y}_n)/n$ converges almost surely, and Kronecker lemma yieldsFinally $\mathbb{E}{Y}_n\to \mu$, so the Cesàro mean of these expectations tends to $\mu$.