Convergent Sequences

What It Means for a Sequence to Converge

A sequence can change at every index and still settle closer and closer to one real number. To make “closer and closer” precise, we specify how close the terms must be and require that all terms from some index onward meet that requirement. This is the central idea of convergence.

In this tutorial, sequences are indexed by $\mathbb{N}_0$. The index after which a condition holds may depend on the required accuracy: demanding a smaller error may require looking farther along the sequence. The definition records this dependence explicitly.

Definition: A real sequence $(a_n)_{n=0}^{\infty}$ converges to $L\in\mathbb{R}$ if, for every $\varepsilon>0$, there exists $N\in\mathbb{N}_0$ such that for every $n\geq N$, $$ |a_n-L|<\varepsilon. $$ In this case, $L$ is called the limit of the sequence, and we write $a_n\to L$ or $\lim_{n\to\infty}a_n=L$. A sequence that converges to some real number is called convergent.

The order of the quantifiers matters. The tolerance $\varepsilon$ is chosen first, and the index $N$ may then depend on it. Once $N$ is chosen, the inequality must hold for every $n\geq N$, not merely for one selected term or for infinitely many terms.

The inequality $|a_n-L|<\varepsilon$ says that $a_n$ lies within distance $\varepsilon$ of $L$. Equivalently, $L-\varepsilon<a_n<L+\varepsilon$. Since the definition must work for every positive tolerance, it forces the terms eventually into arbitrarily small neighborhoods of the proposed limit.

Verifying a Limit from the Definition

A direct proof usually begins with an arbitrary $\varepsilon>0$. The goal is to choose an integer threshold $N$ that makes the error $|a_n-L|$ smaller than $\varepsilon$ whenever $n\geq N$. The useful choice of $N$ is often suggested by simplifying the error first.

Worked Example: A Reciprocal Sequence Converges to Zero

Let $a_n=\dfrac{5}{n+3}$. We prove directly that $a_n\to0$. Let $\varepsilon>0$. Choose $N\in\mathbb{N}_0$ such that $N>\dfrac{5}{\varepsilon}$, which is possible because the nonnegative integers are unbounded above. If $n\geq N$, then $n+3\geq N+3>\dfrac{5}{\varepsilon}>0$, and therefore

|a_n-0|=\frac{5}{n+3}<\varepsilon.

Thus, for each positive tolerance, every term from the chosen index onward is within that tolerance of $0$. By the definition, the sequence converges to $0$.

The threshold does not need to be the smallest possible one. It only needs to be an element of $\mathbb{N}_0$ for which the required inequality holds. This flexibility is useful: a simple bound on the error often gives a convenient choice.

Worked Example: A Rational Sequence Converges to Four

Consider $a_n=\dfrac{4n+1}{n+2}$, where $n\in\mathbb{N}_0$. To test the proposed limit $4$, subtract it from the general term:

|a_n-4| =\left|\frac{4n+1}{n+2}-4\right| =\left|\frac{4n+1-4(n+2)}{n+2}\right| =\frac{7}{n+2}.

Here $n+2>0$, and the numerator inside the absolute value is $4n+1-4n-8=-7$, so the final expression is correct. Given any $\varepsilon>0$, choose $N\in\mathbb{N}_0$ with $N>\dfrac{7}{\varepsilon}$. For $n\geq N$, we have $n+2\geq N+2>\dfrac{7}{\varepsilon}$, hence

|a_n-4|=\frac{7}{n+2}<\varepsilon.

This verifies the definition and proves that $a_n\to4$. The sequence need not equal its limit at any index for convergence to hold; what matters is that the error becomes smaller than every prescribed positive tolerance and stays so thereafter.

A Limit, If It Exists, Is Unique

A sequence cannot converge to two different real numbers. If two proposed limits were separated by a positive distance, then sufficiently small neighborhoods around them would not overlap. But convergence would require the same late terms to lie in both neighborhoods.

Theorem (Uniqueness of Limits): If a real sequence $(a_n)$ converges to $L$ and also converges to $M$, then $L=M$.

Proof. Suppose $a_n\to L$ and $a_n\to M$. Assume, for contradiction, that $L\neq M$. Then $|L-M|>0$, so $\varepsilon=\dfrac{|L-M|}{3}$ is positive. By convergence to $L$, there is an index $N_1$ such that $|a_n-L|<\varepsilon$ whenever $n\geq N_1$. By convergence to $M$, there is an index $N_2$ such that $|a_n-M|<\varepsilon$ whenever $n\geq N_2$.

Let $N$ be the larger of $N_1$ and $N_2$. Then both inequalities hold for $n=N$. By the triangle inequality,

|L-M| =|L-a_N+a_N-M| \leq |L-a_N|+|a_N-M| <2\varepsilon =\frac{2}{3}|L-M|.

This is impossible because $|L-M|>0$, and thus $\frac{2}{3}|L-M|<|L-M|$. The assumption $L\neq M$ must be false, so $L=M$. $\square$

Uniqueness lets us speak of the limit of a convergent sequence. It also gives a useful proof strategy: to establish that a sequence has a particular limit, verify the definition for that number; no competing limit can exist.

Every Convergent Sequence Is Bounded

Convergence controls all sufficiently late terms, but boundedness concerns every term, including the initial ones. The definition handles the tail, and the finitely many terms before the relevant threshold can be included in a bound as well.

Theorem (Convergent Sequences Are Bounded): Every convergent real sequence is bounded.

Proof. Suppose $a_n\to L$. Apply the definition with $\varepsilon=1$. There is an $N\in\mathbb{N}_0$ such that, whenever $n\geq N$,

$$ |a_n-L|<1. $$

The triangle inequality gives $|a_n|\leq |a_n-L|+|L|<1+|L|$ for every $n\geq N$. Thus the tail is bounded in absolute value by $1+|L|$.

There are only finitely many indices $n<N$. If $N=0$, there are no such indices, and $1+|L|$ bounds the whole sequence. If $N>0$, the finite set of numbers $|a_0|,\ldots,|a_{N-1}|$ has a maximum. Let $M$ be the larger of this maximum and $1+|L|$. Then $|a_n|\leq M$ for every index: the finite initial part is covered by its maximum, and the tail is covered by $1+|L|\leq M$. Therefore $(a_n)$ is bounded. $\square$

The converse is not true: boundedness alone does not guarantee convergence. A sequence may stay between fixed bounds while continuing to move between separated values.

Worked Example: A Bounded Sequence That Does Not Converge

Define $a_n=1$ when $n$ is even and $a_n=-1$ when $n$ is odd. Every term has absolute value $1$, so the sequence is bounded. Suppose it converged to some $L$. Using the definition with $\varepsilon=\dfrac12$, there would be an $N$ such that $|a_n-L|<\dfrac12$ for all $n\geq N$. The consecutive indices $N$ and $N+1$ have opposite parity, so one corresponding term equals $1$ and the other equals $-1$. Consequently, both $|1-L|<\dfrac12$ and $|-1-L|<\dfrac12$ would hold. But the triangle inequality would then imply

2=|1-(-1)|\leq |1-L|+|L-(-1)|<1,

a contradiction. Therefore the sequence is bounded but not convergent. This example shows why boundedness and convergence are distinct properties.

Subsequences Preserve the Limit

A subsequence retains terms from the original sequence in increasing index order. If the original terms are eventually close to $L$, then selected terms whose indices are sufficiently large must also be close to $L$. The fact that subsequence indices increase is essential: their indices cannot remain forever among the original sequence’s early terms.

Theorem (Subsequences of a Convergent Sequence): If $a_n\to L$ and $(a_{n_k})_{k=0}^{\infty}$ is a subsequence, then $a_{n_k}\to L$.

Proof. Let $\varepsilon>0$. Since $a_n\to L$, there is an $N\in\mathbb{N}_0$ such that $|a_n-L|<\varepsilon$ for every $n\geq N$. The subsequence indices satisfy $n_k\geq k$, by the lemma on increasing subsequence indices established in “Subsequences.” Thus, whenever $k\geq N$, we have $n_k\geq k\geq N$, and hence

|a_{n_k}-L|<\varepsilon.

This holds for every $k\geq N$. Since the choice works for each positive $\varepsilon$, the subsequence converges to $L$. $\square$

This theorem provides a useful test: if a sequence has two subsequences that converge to different real numbers, the original sequence cannot converge. If it did, both subsequences would have to converge to its unique limit.

Worked Example: A Convergent Sequence and Two Subsequences

For $n\in\mathbb{N}_0$, set $a_n=6+\dfrac{(-1)^n}{n+1}$. Since $|(-1)^n|=1$, the error from $6$ is

|a_n-6|=\frac{1}{n+1}.

Given $\varepsilon>0$, choose $N\in\mathbb{N}_0$ with $N>\dfrac{1}{\varepsilon}$. If $n\geq N$, then $n+1\geq N+1>\dfrac{1}{\varepsilon}$, so $|a_n-6|<\varepsilon$. Therefore $a_n\to6$. The even-indexed terms $a_{2k}$ and the odd-indexed terms $a_{2k+1}$ are both subsequences, with strictly increasing indices. By the subsequence theorem, both converge to $6$, even though their terms approach $6$ from opposite sides.

What the Definition Does—and Does Not—Require

Convergence is an eventual condition: the terms before the threshold $N$ do not affect whether the sequence converges to $L$. Nor does convergence require any term to equal $L$. For example, the terms in the reciprocal example are all positive, yet their limit is $0$.

It is also not enough that some terms get close to $L$ at widely separated indices. The definition requires that every term after one threshold remain within the chosen tolerance. That distinction explains why selecting a subsequence can reveal a limit but cannot disregard terms when proving convergence of the original sequence.

Start with an arbitrary tolerance.
Take an arbitrary $\varepsilon>0$; the proof must work no matter how small it is.

Measure the error.
Simplify $|a_n-L|$ and identify how it depends on $n$.

Choose a threshold.
Find $N$ so that $n\geq N$ makes the error strictly less than $\varepsilon$.

Check the whole tail.
Verify the inequality for every $n\geq N$, then conclude convergence by definition.

These steps separate the guess for a limit from its proof. A pattern may suggest a value, but convergence is established only by meeting the quantifiers in the definition. Once it is established, uniqueness, boundedness, and preservation under taking subsequences follow as proved above.

Check Your Understanding

Use the definition and the results proved here to answer the following questions.

In the definition of convergence, which quantity may depend on the chosen $\varepsilon$, and what must hold for every index after that choice?
Why does the uniqueness proof choose a tolerance based on the distance between two proposed limits?
How does the proof that a convergent sequence is bounded handle the terms before the tail threshold?
Why does an increasing sequence of subsequence indices guarantee that sufficiently late subsequence terms come from the original sequence’s tail?
Can a sequence converge to $L$ without any term being equal to $L$? Explain using the definition or an example.

What You'll Learn