Sequences and Sequence Limits

A sequence is a function on integer indices

An infinite real sequence assigns a real number $a_n$ to each index $n$ in an integer list such as $1,2,3,\ldots$. Some sequences start at $n=0$ or another specified integer. The starting index belongs to the definition.

An explicit rule gives a term directly from its index. For $a_n=n/(n+1)$, starting at $n=1$, the first four terms are $1/2,2/3,3/4,4/5$. An index such as $n=2.5$ is not part of this sequence even though the related real-variable formula can be evaluated there.

A recursive rule uses earlier terms to obtain the next term. It also needs an initial value. A recurrence without a starting value generally does not specify one unique sequence.

Do not confuse the notation $a_n$ with multiplication $a\cdot n$. The subscript identifies which term is being discussed. Changing a finite number of terms does not change whether an infinite sequence converges or the value of its limit, because a limit concerns the eventual tail.

Sequence limits

The statement $a_n\to L$ means: for every tolerance $\varepsilon>0$, there is an integer $N$ such that all terms with $n\ge N$ satisfy

A convergent sequence must eventually remain close to $L$, not merely visit close values repeatedly. This is the discrete limit definition used in the OpenStax sequence chapter.

Prove the limit of a rational sequence

Divide numerator and denominator of $n/(n+1)$ by $n$:

The distance from $1$ gives a more direct check:

For any chosen $\varepsilon>0$, choose a positive integer $N\ge1/\varepsilon$. Then $n\ge N$ makes $1/(n+1)<\varepsilon$. For a tolerance of $0.01$, every term from $n=100$ onward is within that distance of $1$.

The sequence $(1/2)^n$ tends to $0$. A sequence need not be monotone to converge: $(-1)^n/n$ also tends to zero, because its absolute value is $1/n$. The signs alternate while the magnitudes shrink.

By contrast, $(-1)^n$ does not converge. Its even-indexed terms are all $1$, and its odd-indexed terms are all $-1$. These two subsequences cannot approach one common limit. The sequence $n$ diverges in another way: it grows without bound. Divergence does not always mean growth to infinity.

Monotone and bounded sequences

A sequence is non-decreasing when $a_{n+1}\ge a_n$, and non-increasing when $a_{n+1}\le a_n$. Either property is called monotonicity. Strictly increasing or decreasing uses strict inequalities. A sequence is bounded above and below when fixed real numbers contain every term between them.

The monotone convergence theorem states that every bounded monotone real sequence converges. Neither condition is enough alone: $(-1)^n$ is bounded but does not converge, while $a_n=n$ is increasing but unbounded.

For the example $a_n=1-1/n$, with $n\ge1$,

Also, $0\le a_n<1$. These inequalities establish increasing behaviour and boundedness, so the theorem guarantees a finite limit. The explicit formula then identifies that limit as $1$.

The theorem proves that a limit exists; it does not automatically state which bound is the limit. An upper bound need not be the least upper bound. For example, $2$ is also an upper bound for this sequence, but the limit is not $2$.

Recursive sequences

Consider the approximation rule

The first four terms are $1,3/2,17/12,577/408$. A few close decimal values suggest $\sqrt2$, but they are not a proof. We need to show the recurrence stays defined, converges, and has the claimed limit.

Show that the terms remain positive

A positive input gives a positive output, so induction from $a_1=1$ ensures that division by $a_n$ is always valid. For any positive input $s$,

Consequently every term from $a_2$ onward is at least $\sqrt2$.

Show that the tail decreases

For $n\ge2$, the bound $a_n\ge\sqrt2$ gives

The tail is non-increasing and bounded below, so it converges to a limit $L\ge\sqrt2>0$. The first step rises from $1$ to $3/2$; monotonicity begins with the second term, which is enough because a finite prefix does not affect convergence.

The update function is continuous at this positive limit, so passing to the limit is now justified:

Since $L>0$, we obtain $L=\sqrt2$. Solving the fixed-point equation without the positivity and convergence arguments would identify candidates, not establish the sequence's behaviour.

Read a sequence through its values

A second explicit sequence, $b_n=(n+1)/(2n^2)$ for $n\ge1$, shows how a table can support a limit argument. This is the sequence used in Melikamp's open-licence convergence illustration. Its limit is zero, unlike the earlier explicit sequence, whose limit is one.

Writing $b_n=1/(2n)+1/(2n^2)$ proves its limit is zero. The table illustrates the approach; the formula establishes what happens beyond the displayed terms. A small collection of values can suggest a limit, but the argument must cover every sufficiently large integer index.

Sequences are the foundation of series

Given a term sequence $a_n$, define a second sequence of partial sums:

The term $a_N$ is one contribution. The partial sum $S_N$ includes every contribution up to that index. Their limits can differ. For $a_n=(1/2)^n$ with $n\ge1$, the individual terms approach zero while the partial sums approach $1$.

The infinite series $\sum_{n=1}^\infty a_n$ converges precisely when this partial-sum sequence converges to a finite number. Establishing only that $a_n\to0$ is therefore not the same task as establishing series convergence.

Related topics

Continue to Infinite Series and Convergence for geometric sums, telescoping cancellation and examples where terms tend to zero but their partial sums grow without bound.

Practice questions

Sources and further reading

• OpenStax Calculus Volume 2: Sequences. Sequence limits, subsequences, monotonicity and boundedness.
• Wolfram Language: Calculus. Independent symbolic calculations and limit verification.