Convergent sequence

The sequence is bounded and $|a_{n+1}-a_n|\le \sqrt{n+1}-\sqrt{n}\to 0$ by the mean value theorem. Nevertheless, along the integer parts of $(\pi /2+2\pi j{)}^2$ its terms tend to one, and along the integer parts of $(3\pi /2+2\pi j{)}^2$ they tend to minus one. The square root rounding error is $O(1/j)$. Thus a bounded sequence with vanishing successive increments need not be a convergent sequence.

A sequence $(x_n{)}_{n\ge 1}$ of real numbers has limit of a sequence $L\in \mathbb{R}$, written $x_n\to L$, precisely whenThe absolute value measures the error. The tolerance $\epsilon$ is chosen first; $N$ may depend on it, but must control every later term. Infinitely many accurate terms alone do not suffice. For example, $x_n=(-1{)}^n$ repeatedly equals one but also repeatedly differs from one by two. The definition and uniqueness of the limit of a sequence are discussed in MIT 18.100A, Lecture 6, definitions 13 and theorem 15.

For a worked error estimate, let $x_n=3+2/n$. Then $|x_n-3|=2/n$, so any positive integer $N>2/\epsilon$ works. With $\epsilon =0.01$, choose $N=201$; every later error is at most $2/201<0.01$. The strict inequality matters: $N=200$ would give equality at its first term. A limit of a sequence turns the informal instruction “take enough terms” into an explicit accuracy guarantee.

The limit of a sequence is unique: two distinct proposed answers cannot both be within arbitrarily small error of every late term, by the triangle inequality. Changing finitely many initial terms does not change it. For an extended real number, $x_n\to +\infty$ means $\forall M\in \mathbb{R}\exists N\forall n\ge N:x_n>M$; convergence to $-\infty$ similarly requires $x_n<M$. These describe escape rather than a finite answer. The extended-real Fekete lemma is an example where the limit of a sequence need not be finite.

For a real sequence with no zero terms, the displayed equivalence follows directly by comparing $|1/a_n|$ with $1/M$. Escape to positive infinity implies reciprocal convergence to zero, but the converse does not retain the sign: $a_n=-n$ is a counterexample. If $a_n\to a\ne 0$, eventually $|a_n|>|a|/2$, and $|1/a_n-1/a|\le 2|a_n-a|/|a{|}^2$. This proves reciprocal convergence at a nonzero finite limit of a sequence.

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.

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.

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.

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 positive sequence converging to $L\ge 0$, its cumulative geometric means converge to the same limit.

An increasing real sequence converges exactly when it is bounded above. For a bounded sequence, its limit is the supremum of its set of terms.