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If a sequence converges, its Cesaro means have the same limit. Given a small tolerance, split the average error into a fixed finite initial segment and a tail whose individual errors are below that tolerance. The first contribution is a fixed constant divided by $n$, while the second is bounded by the tolerance. The converse fails: the alternating sequence $(-1{)}^n$ has means tending to zero while its terms do not converge.

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 Cesaro means $m_N$ converge.

Let $d_j=a_{j+1}-a_j$ and suppose $d_j\to L$. Telescoping givesTo justify the Cesaro 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 Cesaro 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 Cesaro 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 Cesaro mean of these expectations tends to $\mu$.

If $x_n\rightharpoonup x$ in a Hilbert space, some subsequence $(x_{n_k})$ has Cesaro means converging in norm to $x$. For $x=0$, choose the subsequence so successive vectors have summably small inner products; expansion of the squared norm of its averages then proves convergence.