Sequences, Recurrence, and Convergence

Define sequences explicitly or recursively, compare arithmetic and geometric change, and sum finite geometric series. Use limits and fixed points to distinguish observed updates from justified convergence claims.

A sequence assigns a value to each index in an ordered set, usually consecutive integers. An explicit rule computes a term directly, while a recurrence rule computes new terms from earlier ones. Studying how those terms behave reveals growth, decay, oscillation, and convergence.

A Sequence Is a Function of an Index

The sequence

2,\ 5,\ 8,\ 11,\ldots

can be defined by:

a_n=2+3n, \qquad n=0,1,2,\ldots.

The symbol $a_n$ denotes the term at index $n$ . This is an explicit rule: the index alone is enough to compute any term directly. For example:

a_4=2+3(4)=14.

The index set is part of the definition. Because this sequence begins at $n=0$ , the term $a_4$ is the fifth displayed term. If the same visible list were indexed from $n=1$ , one suitable rule would be $a_n=3n-1$ .

Exercise: Evaluate an explicit sequence

Let $a_n=4-2n$ for $n=0,1,2,\ldots$ . What is $a_5$ ?

Your answer

Compute it first, then check your number.

HintUse the index as the input

Compute $4-2(5)$ .

SolutionSubstitute the requested index

Substitution gives:

a_5=4-2(5)=4-10=-6.

Since indexing starts at $0$ , this is the sixth displayed term. The subscript identifies a position; it does not multiply $a$ by $5$ .

A Recurrence Defines Transitions between Terms

The same arithmetic sequence can be defined recursively:

a_0=2, \qquad a_{n+1}=a_n+3.

The first equation supplies an initial condition. The second gives the update from the current term to the next one. Both parts are needed: the update rule alone describes many sequences with different starting values.

A recurrence may depend on more than one earlier term. For example, the Fibonacci recurrence

F_{n+1}=F_n+F_{n-1}, \qquad n\ge1,

needs two initial values, such as $F_0=0$ and $F_1=1$ , before it determines a unique sequence.

Repeated updates in finance, population models, numerical algorithms, and machine learning often have recurrence form. A parameter update, for example, may be written:

w_{t+1}=w_t-\eta g_t.

This equation defines one transition from the current parameter $w_t$ to the next parameter. It does not state the final value directly.

Exercise: Trace repeated updates

Let $w_0=10$ and $w_{t+1}=w_t-2$ . What is $w_3$ ?

Your answer

Compute it first, then check your number.

HintWrite each state

Compute $w_1$ , then $w_2$ , then $w_3$ .

SolutionApply the recurrence step by step

Starting from $w_0=10$ :

w_1=8, \qquad w_2=6, \qquad w_3=4.

Reaching $w_3$ requires three transitions. The subscript counts update steps; it is not multiplication.

Arithmetic and Geometric Rules Produce Different Change

An arithmetic sequence adds a fixed difference $d$ :

a_{n+1}=a_n+d.

If it starts at $a_0$ , its explicit form is:

a_n=a_0+nd.

The change is additive: every step has the same difference.

A geometric sequence multiplies by a fixed ratio $r$ :

b_{n+1}=r b_n.

If it starts at $b_0$ , its explicit form is:

b_n=b_0r^n.

For $b_0=3$ and $r=\frac12$ , the terms are:

3,\ \frac32,\ \frac34,\ \frac38,\ldots.

Arithmetic sequences change linearly with the index. Geometric sequences change exponentially because each new step multiplies the accumulated result.

Exercise: Distinguish a ratio from a difference

The sequence $5,10,20,40,\ldots$ is geometric. What is its common ratio?

Your answer

Compute it first, then check your number.

HintCompare consecutive terms multiplicatively

Compute $10/5$ , $20/10$ , and $40/20$ .

SolutionDivide consecutive terms

Each consecutive ratio is the same:

\frac{10}{5}=\frac{20}{10}=\frac{40}{20}=2.

Therefore $r=2$ . The differences are not constant, so the sequence is not arithmetic.

A Series Adds the Terms of a Sequence

A series is a sum of sequence terms. The finite geometric series

S_n=1+r+r^2+\cdots+r^n

contains $n+1$ terms. For $r\ne1$ , multiply by $r$ and subtract:

rS_n=r+r^2+\cdots+r^{n+1},

so:

S_n-rS_n=1-r^{n+1}.

Factoring the left side and dividing by $1-r$ gives the closed form:

S_n=\frac{1-r^{n+1}}{1-r}, \qquad r\ne1.

When $r=1$ , every term equals $1$ , so $S_n=n+1$ instead.

For $r=\frac12$ and $n=3$ :

1+\frac12+\frac14+\frac18=\frac{15}{8}.

The formula gives the same result:

\frac{1-(1/2)^4}{1-1/2} =\frac{1-1/16}{1/2} =\frac{15}{8}.

Exercise: Evaluate a finite geometric series

What is $1+\frac13+\frac19$ ?

Choose the sum

13 / 9

5 / 3

4 / 3

Select one choice, then check.

HintUse a common denominator

Rewrite $1$ as $9/9$ and $1/3$ as $3/9$ .

SolutionAdd the three terms

Using denominator $9$ :

1+\frac13+\frac19 =\frac99+\frac39+\frac19 =\frac{13}{9}.

This is the finite geometric sum with ratio $1/3$ and indices from $0$ through $2$ .

Convergence Describes Eventual Closeness

A sequence converges to a limit $L$ if its terms become and remain as close to $L$ as desired after a sufficiently large index. More precisely, for every tolerance $\varepsilon>0$ , there is an index $N$ such that:

n\ge N \quad\Longrightarrow\quad |a_n-L|<\varepsilon.

For every positive tolerance epsilon, there is an index $N$ such that every term with index at least $N$ lies within epsilon of $L$ .

We then write:

\lim_{n\to\infty}a_n=L.

For:

b_n=\left(\frac12\right)^n,

the terms

1,\ \frac12,\ \frac14,\ \frac18,\ldots

converge to $0$ . Every finite term remains positive, so convergence does not mean that the sequence must ever equal its limit.

Exercise: Distinguish approaching from reaching

For $b_n=(1/2)^n$ , which statement is correct?

Choose the long-run description

The terms remain positive and approach 0 as n increases.

The sequence becomes exactly 0 after enough finite steps.

The terms grow because the exponent increases.

Select one choice, then check.

HintFollow the multiplier

Each term is half of the preceding positive term.

SolutionDescribe the limiting behavior

Every finite term $(1/2)^n$ is positive. Repeated multiplication by $1/2$ makes the terms arbitrarily close to $0$ , so:

\lim_{n\to\infty}\left(\frac12\right)^n=0.

The sequence approaches its limit without reaching it at a finite index.

For a geometric sequence $b_0r^n$ , the ratio determines the long-run pattern:

if $|r|<1$ , the sequence converges to $0$ ;
if $r=1$ , it remains constant at $b_0$ ;
if $r=-1$ and $b_0\ne0$ , it alternates and does not converge;
if $|r|>1$ and $b_0\ne0$ , its magnitude grows without bound.

When $|r|<1$ , the infinite geometric series also converges:

1+r+r^2+\cdots=\frac{1}{1-r}.

This is a statement about the limit of its finite partial sums, not an instruction to finish infinitely many additions.

Fixed Points Help Analyze Recurrences

Suppose a recurrence has the form $x_{n+1}=F(x_n)$ . A fixed point is a value $L$ that remains unchanged by the update:

F(L)=L.

If the sequence converges to $L$ and $F$ behaves continuously near $L$ , then the limit must satisfy this fixed-point equation. The equation identifies a candidate limit; it does not by itself prove that the sequence converges to that candidate.

For example:

x_{n+1}=\frac{x_n+6}{2}

has the fixed point found from:

L=\frac{L+6}{2} \quad\Longrightarrow\quad 2L=L+6 \quad\Longrightarrow\quad L=6.

Starting from $x_0=2$ gives $2,4,5,5.5,\ldots$ , which suggests convergence toward $6$ . A complete proof would still need to justify the long-run claim.

Exercise: Find a fixed point

What is the fixed point of $x_{n+1}=0.5x_n+3$ ?

Your answer

Compute it first, then check your number.

HintSet the update equal to its input

Solve $L=0.5L+3$ .

SolutionSolve the fixed-point equation

A fixed point remains unchanged, so:

L=0.5L+3.

Subtracting $0.5L$ from both sides gives $0.5L=3$ , hence $L=6$ . This identifies the fixed point; convergence from a particular initial value is a separate question.

Trace Finite Steps before Claiming a Limit

When reading a recurrence, identify its initial conditions and update rule, then compute several indexed terms without skipping transitions. Look for constant differences or ratios, fixed points, cycles, growth, and decay. These finite observations can suggest long-run behavior, but a convergence claim must say what limit is approached and why the terms eventually remain arbitrarily close to it.