Sequence and series

A sequence is an ordered family indexed by the natural numbers; a series studies the partial sums of a sequence of terms.

A product in which numerator and denominator factors cancel between successive terms. For nonzero $b_m,\dots ,b_{n+1}$,One can prove the identity directly by cancellation or by mathematical induction on $n$. Factoring a more complicated product into two such ratios often reveals an otherwise hidden closed expression.

If $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. The nondecreasing partial sums of the nonnegative $a_n$ are bounded above by a finite initial contribution plus the sum of $b_n$; bounded monotone sequence theorem proves the claim. Consequently $|c_n|\le b_n$ with summable $b_n$ proves absolute convergence of $\sum c_n$. Finite changes to initial terms do not affect convergence.

For $0\le r<2^n$ and $n\ge 0$, the displayed sequence tends to zero, but its series diverges. The signed sum across the $n$th dyadic block has magnitude ${\sum }_{r=0}^{2^n-1}(2^n+r{)}^{-1}\ge 1/2$. Differences of arbitrarily late partial sums therefore fail the Cauchy sequence condition on partial sums. Alternating signs between blocks do not rescue convergence when the total contribution of an entire block stays away from zero.

The Lucas numbers obey the same second-order recurrence as the Fibonacci numbers but start with $2,1$, so their sequence is $2,1,3,4,7,11,\dots$. For $n\ge 1$, $L_n=F_{n-1}+F_{n+1}$. Substitution in the Fibonacci addition formula gives $F_{2n}=F_n{L}_n$. Consecutive Fibonacci numbers are coprime by the Euclidean algorithm; since $L_n=F_n+2F_{n-1}$, Bézout's identity gives $\gcd (F_n,L_n)=\gcd (F_n,2)\le 2$.

For $\phi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2$, each power sequence satisfies the recurrence relation $x_{n+2}=x_{n+1}+x_n$, because both bases solve $t^2=t+1$. Their sum starts at $2,1$, so uniqueness of a recursively defined sequence proves that it is the Lucas number sequence. Alternatively, diagonalize the companion matrix $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ and use $(1,2{)}^T=(\phi ,1{)}^T+(\psi ,1{)}^T$. The formula explains the exponential growth governed by the golden ratio and the alternating decaying correction ${\psi }^n$.

A sequence assigns a value to each position in an ordered index list. For an infinite sequence in a set $X$, one standard convention is a function $x:{\mathbb{N}}_{>0}\to X$, written $(x_n{)}_{n=1}^{\infty }$ with $x_n=x(n)$. Some applications start the index at zero, and a finite sequence has only finitely many positions. Values may repeat; both order and repetition matter. ^[ref].

For example, $a_n=(-1{)}^n$ begins $-1,1,-1,1,\dots$, whereas $b_n=(-1{)}^{n+1}$ begins $1,-1,1,-1,\dots$. Their sets of values are both $\{-1,1\}$, but these are different sequences. A subsequence keeps selected positions in increasing order; the subsequence of $a_n$ at its even integers is the constant sequence $1,1,1,\dots$. A tail of a sequence retains all positions from one index onward.

Sequences express successive approximations, repeated observations, and the iterates of a recurrence relation. A limit of a sequence describes where its terms eventually concentrate. Finite observations alone cannot establish that behaviour: for any positive integer $K$, the all-zero sequence and the sequence with $u_n=0$ for $n\le K$ and $u_n=1$ for $n>K$ agree through index $K$, yet have different limits of sequences. A formula or bound controlling the entire future, such as the tail diameter of a sequence in the Cauchy sequence condition, supplies information that a finite list cannot.

A sequence formed by an ascending part followed by a descending part, or the reverse arrangement, with either part possibly empty. In sorting algorithms, the monotonicity convention normally allows equal keys. Joining one ascending sequence to the reversal of another ascending sequence produces a bitonic sequence.

A positive sequence is superincreasing if each term exceeds the sum of all earlier terms. If $w_{k+1}>2w_k$, a geometric series proves ${\sum }_{j<k}{w}_j<w_k$. Distinct finite subsets of its indices have distinct sums: cancel their common terms and compare the largest remaining term with the sum of all smaller terms. This gives uniqueness of minimum spanning trees and of simple graph paths with fixed endpoints when a finite weighted graph has such separated edge weights.

A sequence equal to its own sequence of consecutive run lengths. The classical sequence uses the alphabet $\{1,2\}$ and begins $1,2,2,1,1,2,1,2,2,\dots$; variants use other pairs of positive integers.

A sequence is eventually periodic if there are integers $N\ge 0$ and $T>0$ such that $u_{n+T}=u_n$ for all $n\ge N$. It consists of a finite initial segment followed by a repeating period. Checking finitely many repeated values can suggest eventual periodicity, but a proof requires a reason the repetition continues, such as recurrence of a complete state in a deterministic finite-state process.

Suppose a numerical sequence has period $T$ from index $N$, and let $C={\sum }_{j=0}^{T-1}{u}_{N+j}$. For $L=qT+r$ with $0\le r<T$, split the required sum into $q$ complete periods and a tail of length $r$. Every complete period contributes $C$, proving the formula in the title. Precomputed prefix sums of one period make the tail lookup constant time. If this periodic sequence consists of the first differences of another sequence, the same formula extrapolates that original sequence by adding the initial value at $N$.

Each term is obtained by reading the lengths and values of consecutive runs in the previous term. Starting from $1$ gives $1,11,21,1211,111221,\dots$: for example, $21$ contains one $2$ followed by one $1$, so its successor is $1211$.

For a nonnegative integer $n$, let $r(n)$ count overlapping occurrences of $11$ in its binary expansion. The Rudin-Shapiro sign sequence is $b(n)=(-1{)}^{r(n)}$. Appending a binary digit gives $b(2n)=b(n)$ and $b(2n+1)=(-1{)}^{n}b(n)$.

The sequence $\lfloor n\alpha \rfloor$ for $n=1,2,\dots$, with a fixed positive irrational number $\alpha$. Complementary Beatty sequences for $\alpha ,\beta >1$ partition the positive integers when $1/\alpha +1/\beta =1$.

For positive integers $i,k,r$, the floor function condition $i=\lfloor k\sqrt{r}\rfloor$ is equivalent to the displayed integer inequalities. Since $r\ge 1$, a witness satisfies $k\le i$. Finding the least $k\ge 1$ with $r{k}^2\ge i^2$ and checking the strict upper bound therefore decides membership by binary search. Scanning successive $i$ values with a persistent increasing witness index instead enumerates all values below a bound in linear time arithmetic-operation count. Exact integer comparison avoids floating-point errors near an integer boundary.

If $\alpha ,\beta >1$ are irrational numbers with $1/\alpha +1/\beta =1$, the Beatty sequences $\lfloor n\alpha \rfloor$ and $\lfloor n\beta \rfloor$ partition the positive integers.

A word formed recursively by concatenating the previous two words in a specified order, for example $W_n=W_{n-2}{W}_{n-1}$ from two seed words. Lengths satisfy the Fibonacci recurrence. To locate a symbol, compare its index with the first block's length, choose the containing block, and subtract the first length when entering the second block. This avoids constructing an exponentially long word.

The integer sequence proposed by Bernardo Recamán Santos starts at zero. At step $n>0$, subtract $n$ if the result is positive and has not appeared before; otherwise add $n$. The commonly written nonnegative version gives the same sequence because zero is already present. The initial values are $0,1,3,6,2,7,13,20,\dots$. The addition branch can repeat a value: $r(20)=r(24)=42$. Avoiding a repeated subtraction is therefore not a global injectivity condition.^[ref]

Since $r(n)\le r(n-1)+n$, induction gives $0\le r(n)\le n(n+1)/2$. A prefix scan needs only constant extra storage but can take quadratic membership work. A bitset-style bounded membership table makes each lookup constant time, while its allocation cost depends on the chosen value universe. This separates the number of generated terms from the range of possible stored values.

A divisibility sequence associated with multiples of a point on an elliptic curve, using suitably normalized division polynomials. With $W_0=0$ and $W_1=1$, its terms satisfy the Ward identity $W_{m+n}{W}_{m-n}=W_{m+1}{W}_{m-1}{W}_n^2-W_{n+1}{W}_{n-1}{W}_m^2$. Setting nearby indices gives doubling identities, enabling logarithmic-depth evaluation.

Starting with two distinct positive integers $a<b$, repeatedly append the smallest integer larger than the previous term that is expressible in exactly one way as the sum of two distinct earlier terms. Requiring distinct summands and counting unordered pairs are both essential parts of the definition.

The nondecreasing sequence of positive integers beginning with $G(1)=1$ in which the integer $n$ occurs exactly $G(n)$ times. Its first terms are $1,2,2,3,3,4,4,4,5,5,5,\dots$. The prefix sum ${\sum }_{j=1}^{k}G(j)$ gives the last position occupied by $k$.

Terms of a sequence satisfying the third-order linear recurrence $T_{n+3}=T_n+T_{n+1}+T_{n+2}$. Initial values vary by convention; some counting problems specify $T_1=T_2=T_3=1$. To study divisibility by $m$, retain three consecutive remainders modulo $m$, a deterministic state with at most $m^3$ possibilities.

For a positive integer $a_0$, define $a_{k+1}=a_k/2$ when $a_k$ is an even number, and $a_{k+1}=3a_k+1$ when $a_k$ is an odd number. Starting at $6$ gives $6,3,10,5,16,8,4,2,1$. Computing many trajectories benefits from memoization, since trajectories that reach the same value have identical continuations.

A constant sequence has the same value $a$ at every index. It converges to $a$ by convergence of a sequence in a topological space.

A sequence indexed from zero with an integer period $p\ge 1$ satisfying the displayed identity for all $n\ge 0$. Repeating a finite cycle is an example. The Cesàro mean article explains why its cumulative averages approach the one-cycle arithmetic mean.

A tail of a sequence consists of its terms from some index $N$ onward, retaining their order and repetitions. Discarding finitely many initial terms does not change its limit of a sequence or whether it is a Cauchy sequence. A tail can also be considered as the set of values $\{x_n:n\ge N\}$ when forming an infimum, supremum, or tail diameter of a sequence; that set alone forgets the original order and repetitions.

The limit inferior and limit superior retain the lower and upper long-run behaviour of a real sequence, even when it has no ordinary limit of a sequence. For each tail of a sequence beginning at $N$, defineDiscarding more initial terms raises or preserves the infimum and lowers or preserves the supremum. Hence $(l_N)$ is nondecreasing and $(u_N)$ is nonincreasing. Their limits exist in the extended real number line:For a bounded sequence, both values are finite and $\mathrm{liminf}{a}_n\le \mathrm{limsup}{a}_n$. These bounds use the entire infinite tail of a sequence, rather than the last few observed terms. ^[ref].

Take the original example $a_n=(-1{)}^n(1+1/n)$, beginning $-2,3/2,-4/3,5/4,\dots$. The subsequence at even integers decreases toward one, while the subsequence at odd integers increases toward minus one. If $e_N$ is the first even integer at least $N$ and $o_N$ is the first odd integer at least $N$, thenThus $\mathrm{liminf}{a}_n=-1$ and $\mathrm{limsup}{a}_n=1$. The finite initial extreme $a_1=-2$ disappears from later lower bounds; neither long-run value is attained by any term of this example.

For a bounded sequence, the limit inferior is the smallest subsequential limit and the limit superior is the largest subsequential limit; both are reached by suitable convergent subsequences. Every other subsequential limit lies between them. In the example there are only two subsequential limits, $-1$ and $1$: a convergent subsequence cannot retain infinitely many terms of both parities, because those two branches stay separated. The whole interval between the two bounds need not consist of subsequential limits. ^[ref].

A finite ordinary limit of a sequence exists exactly when the limit inferior and limit superior agree at the same finite value $L$. The bounds $l_N\le a_N\le u_N$ and the squeeze theorem prove the sufficient direction. For a bounded sequence, its tail diameter of a sequence is $D_N=u_N-l_N$, so the limiting gap is zero exactly when the Cauchy sequence condition holds. In the example the gap approaches two, not zero. The same framework allows infinite bounds: $a_n=n$ has both values $+\infty$, whereas $a_n=(-1{)}^{n}n$ has limit inferior $-\infty$ and limit superior $+\infty$. Equality at infinity describes escape rather than convergence to a real value. ^[ref].

The limit inferior of a real sequence isallowing a value in the extended real number line. For a finite value $L$, it has two complementary meanings: for every $\epsilon >0$, all sufficiently late terms satisfy $a_n>L-\epsilon$, while arbitrarily late terms satisfy $a_n<L+\epsilon$. The first prevents persistent dips below $L$; the second prevents any larger number from being the true lower edge.

For $a_n=(-1{)}^n(1+1/n)$, the lower bounds in limit inferior and limit superior approach $-1$. Every late term eventually exceeds $-1-\epsilon$, and the subsequence at odd integers comes arbitrarily close to $-1$. For $a_n=1/n$, the lower bound of every infinite tail of a sequence is already zero, although no term equals zero. Thus a limit inferior need not be the minimum of any tail.

A recurrence relation specifies members of a sequence in terms of previous members, together with enough initial values to determine the sequence. A three-term recurrence for orthogonal polynomials relates consecutive orders. For example, the physicists' Hermite polynomials satisfy $H_{n+1}(x)=2x{H}_n(x)-2n{H}_{n-1}(x)$, starting from $H_0=1$ and $H_1=2x$.

Prove a proposed bound by mathematical induction, substituting the inductive hypothesis for smaller arguments into the recurrence relation. Choose its constants so the resulting inequality implies the proposed bound at the current argument, and verify all base cases. A leading-order guess may need an additional lower-order term to make the induction close. The entropy potential for a divide-and-conquer recurrence is an example using a linear correction to an $n\log n$ bound.

If $S_0=a$ and $S_n=S_{\lfloor n/2\rfloor }+m$ for positive integers $n$, each halving removes one binary digit until zero is reached. There are $1+\lfloor {\log }_{2}n\rfloor$ such steps. The formula follows by strong induction, using $\lfloor {\log }_2\lfloor n/2\rfloor \rfloor =\lfloor {\log }_{2}n\rfloor -1$ for $n\ge 2$ and checking $n=1$ separately. With fixed positive $m$, this is $\Theta (\log n)$ as $n\to \infty$; the zero case retains its separately defined value.

For $T(n)=1$ at $n\le 3$ and $T(n)=2T(\lfloor n/2\rfloor {)}^2$ otherwise, put $d=\max (0,\lfloor {\log }_{2}n\rfloor -1)$ for positive integer $n$. Taking logarithms gives $S(n)=1+2S(\lfloor n/2\rfloor )$, hence $S(n)=2^d-1$ and $T(n)=2^{2^d-1}$. These constant-value plateaux have exponential envelopes $2^{\Theta (n)}$, but do not justify $\Theta (2^{n/2})$ for all integer inputs.

For divide-and-conquer recurrences $T(n)=aT(n/b)+f(n)$ with $a\ge 1$, $b>1$, compare $f(n)$ with $n^{{\log }_{b}a}$. If $f(n)=O(n^{{\log }_{b}a-\epsilon })$ for some $\epsilon >0$, then $T(n)=\Theta (n^{{\log }_{b}a})$. If $f(n)=\Theta (n^{{\log }_{b}a})$, then $T(n)=\Theta (n^{{\log }_{b}a}\log n)$. If $f(n)=\Omega (n^{{\log }_{b}a+\epsilon })$ and $af(n/b)\le cf(n)$ for a constant $c<1$ eventually, then $T(n)=\Theta (f(n))$. Constant-size base cases and ordinary bounded rounding of subproblem sizes are understood. A recursion tree proves these cases by summing per-level costs.

For $0<\alpha <1$, the successive differences satisfy $d_{n+1}=-(1-\alpha )d_n$. Summing the resulting geometric series givesEquivalently, the two-dimensional update matrix has eigenvalues $1$ and $\alpha -1$, with only the second mode decaying. At $\alpha =0$, generic initial values instead alternate and do not converge.

A linear recurrence ${\sum }_{j=0}^r{p}_j(n)a_{n+j}=0$ whose coefficients $p_j(n)$ are polynomials in the index. A sequence satisfying such a relation is P-recursive.

A difference equation relates values of a sequence at different indices, or its finite differences. A first-order autonomous system has the displayed form, with a scalar or vector state. Writing $z_{n+1}-z_n=F(z_n)-z_n$ gives its equivalent finite-difference form. Its fixed point stability is determined by the Jacobian matrix of the update and the unit circle, whereas a continuous-time differential equation uses the real parts of its linearized eigenvalues.

Here $s_2(n)$ is the sum of the binary digits of the nonnegative integer $n$. The sign version is $f(n)=(-1{)}^{s_2(n)}$. Concatenating binary digits gives $f(a{2}^j+r)=f(a)f(r)$ for $0\le r<2^j$. It supplies cancellation controlled by binary blocks rather than multiplicative factorization.

Each binary digit independently contributes either $1$ or $-e(2^l\theta )$, proving the product. Two adjacent factors have modulus $8|\cos (\pi t)|(1-{\cos }^2(\pi t))\le 16/(3\sqrt{3})<4$. Consequently the product is $O(2^{\sigma j})$, with $\sigma =2-\frac{3}{4}{\log }_{2}3<1$. This two-factor estimate suffices for a power saving, though its exponent is not optimal.

An integer interval is a disjoint union of aligned binary blocks with at most two of each length. On $[a{2}^j,(a+1)2^j)$ the Thue–Morse sequence sign factors as $f(a)f(r)$, so the Thue–Morse Fourier product bounds its exponential sum by $O(2^{\sigma j})$. Sum the geometric series over block lengths to obtain the uniform bound, for $\sigma =2-\frac{3}{4}{\log }_{2}3$.

The additive character orthogonality filter $1_{m\mid r}=m^{-1}{\sum }_{a=0}^{m-1}e(ar/m)$ restricts the uniform interval cancellation for binary digit parity to multiples of $m$, without increasing its bound. Each inner Type I sum is therefore $O(N^{\sigma })$. Summing the at most $N^{1/100}$ outer terms proves the displayed estimate, which is smaller than every fixed negative power of $\log N$ for large $N$.

Two bounded real sequences with the same set of subsequential limits admit permutations that match their terms with differences tending to zero. Alternate between taking the least unused index of the first sequence and of the second. Match its value to an unused term of the other sequence near a nearest common cluster point, with additional error at most $1/j$ at stage $j$. Every cluster-point neighbourhood contains infinitely many available terms. The actively chosen indices tend to infinity, whose distance to the common cluster set tends to zero by boundedness. By step $2j$, both lists have included every index up to $j$.

A sequence is Fejér monotone with respect to a nonempty set $C$ if $\Vert z^{k+1}-p\Vert \le \Vert z^k-p\Vert$ for every $p\in C$. It is bounded. If it has a cluster point $\bar{z}\in C$, its nonincreasing distance to $\bar{z}$ and that convergent subsequence force the entire sequence to converge to $\bar{z}$.

A finite repetition-free sequence in a set $X$ is an injective function from an initial segment $\{0,\dots ,k-1\}$ of the natural numbers into $X$, including the empty sequence for $k=0$. For a finite set of size $m$, the number of these sequences isThe order of each list is part of the data, so it can be enumerated without choosing an order on its underlying support.

A real sequence $(a_n)$ is subadditive when $a_{m+n}\le a_m+a_n$ for all positive $m,n$. Fekete lemma givesallowing the value negative infinity.

If the real error depends only on the second index and ${\alpha }_n/n\to 0$, the normalized sequence has a limit in $[-\infty ,\infty )$. For fixed $k$, write $n=qk+r$ with $1\le r\le k$ and iterate to obtain $x_n\le x_r+q(x_k+{\alpha }_k)$. It follows that $\mathrm{limsup}{x}_n/n\le (x_k+{\alpha }_k)/k$ for every $k$; a subsequence attaining the lower limit proves convergence. The limit is ${\inf }_k(x_k+{\alpha }_k)/k$, even when the errors can be negative.

A geometric progression is a sequence obtained by repeatedly multiplying by one fixed common ratio $r$:Here $a$ and $r$ may be real numbers or complex numbers. The recurrence is meaningful even when $a=0$ or $r=0$; expressing $r$ as a ratio of successive terms requires a nonzero denominator. As a polynomial convention, $r^0=1$, including at $r=0$.

For example, quantities $100,80,64,51.2,\dots$ retain four fifths of their preceding value at every step. The decrement changes, but the retained fraction does not. With the same initial value and common ratio $2$, the terms are $100,200,400,800,\dots$. This multiplicative structure is useful whenever the same proportional rule repeats: it turns many updates into the single exponentiation $r^n$. An arithmetic progression instead repeats an additive increment.

For nonzero $a$, the modulus of $a_n$ tends to zero when $|r|<1$, stays fixed when $|r|=1$, and grows without bound when $|r|>1$. For a negative real common ratio, successive nonzero terms also alternate in sign. Thus the terms $1,-1/2,1/4,-1/8,\dots$ decay in size while crossing zero at every step. Adding the terms gives a geometric series; the individual terms and their accumulated partial sums are different sequences.

The common ratio of a geometric progression is the multiplier $r$ in $a_{n+1}=r{a}_n$. Whenever $a_n\ne 0$, it can be recovered as $a_{n+1}/a_n$. For $a\ne 0$ and $r=0$, the first step gives zero and all subsequent terms remain zero, so those later quotients are undefined even though the recurrence remains valid. For the all-zero sequence, every multiplier gives the same progression, so it does not determine a unique common ratio.

A subsequence of $(a_n)$ has the form $(a_{n_k})$, where $n_1<n_2<\cdots$ is a strictly increasing sequence of indices.

A monotone subsequence is a subsequence which is a monotone sequence. The indices always increase strictly, but equal selected values are permitted. The monotone subsequence theorem guarantees one for every sequence of real numbers.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.

Every real sequence has a monotone subsequence, without a boundedness assumption. Call an index a peak when its term is at least every later term. Infinitely many peaks give a nonincreasing subsequence; beyond finitely many peaks each term has a larger later term, which recursively gives an increasing subsequence.

For a sequence of real numbers, a common finite limit of a sequence for all monotone subsequences forces the whole sequence to have that limit of a sequence. There is no need to assume a bounded sequence if the hypothesis already includes convergence of every monotone subsequence.

If $a_n$ fails to tend to $L$, select a subsequence with $|a_{n_j}-L|\ge \epsilon$ for some fixed $\epsilon >0$. The monotone subsequence theorem gives a monotone subsequence of this subsequence, which contradicts the common limit of a sequence $L$. Conversely, every subsequence of a convergent sequence has the original limit of a sequence.

For a bounded sequence, the bounded monotone sequence theorem already ensures that all monotone subsequences converge. Only agreement of their limits of sequences remains to be checked.

A peak of a sequence of real numbers is an index $k$ whose value is at least every later value. This is a condition on the entire tail, stronger than comparing only adjacent terms. Equality is allowed: every index of a constant sequence is a peak of a sequence.

Infinitely many peaks of a sequence, listed in increasing order of index, give a nonincreasing monotone subsequence. If only finitely many exist, every index beyond the last has a larger later term; repeatedly choosing one gives a strictly increasing monotone subsequence. This proves the monotone subsequence theorem.

A peak of a sequence of $(-a_n)$ supplies a lower bound for all subsequent terms of $(a_n)$. If both $(a_n)$ and $(-a_n)$ have a peak of a sequence, the two selected values bound a common tail. Adding bounds for the finite initial segment proves that $(a_n)$ is a bounded sequence.

Suppose $k_1<k_2<\cdots$ are peaks of a sequence $(a_n)$ and $m_1<m_2<\cdots$ are peaks of a sequence $(-a_n)$. Then $(a_n)$ converges precisely when these two subsequences have a common finite limit of a sequence $L$.

Necessity follows because every subsequence inherits the limit of a sequence of a convergent sequence. For sufficiency, choose one selected value from each subsequence within $\epsilon$ of $L$. Every term after both selected indices lies between the two selected values:This is exactly $a_n\to L$.

Suppose both $(a_n)$ and $(-a_n)$ have only finitely many peaks of a sequence. Beyond some index $N$, every term has a strictly larger later term and a strictly smaller later term. Starting from $a_N$, construct one strictly increasing and one strictly decreasing monotone subsequence.