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Sequences · Tutorial 152 of 1000

Notation for Sequences

Read standard sequence notation precisely, including indices, terms, and tails, and use it to compare sequences and their boundedness.

Intermediate 9 min read

What You'll Learn

  • Interpret \(a_n\), \((a_n)_{n=0}^{\infty}\), and index-range notation
  • Distinguish a sequence from an individual term and from its range
  • Read ellipses and identify the indices of displayed terms
  • Define a shifted tail and relate tail equality to eventual agreement
  • Prove that boundedness is unchanged by deleting finitely many initial terms

Reading Sequence Notation

In the previous tutorial, a real sequence was defined as a function from \(\mathbb{N}_0\) to \(\mathbb{R}\), with the value at index \(n\) written \(a_n=a(n)\). In practice, sequences are often displayed using subscripts rather than function notation. This is convenient, but the subscripts carry important information: they identify which term is being named and which indices are included.

The notation \((a_n)_{n=0}^{\infty}\) means the sequence whose term at each nonnegative integer \(n\) is \(a_n\). It is another way of referring to the function \(a:\mathbb{N}_0\to\mathbb{R}\). The subscript \(n=0\) specifies the starting index, and the superscript \(\infty\) indicates that the indices continue without a final one. When the starting index is already clear, the shorter notation \((a_n)\) is common.

Definition: If \(a:\mathbb{N}_0\to\mathbb{R}\) is a real sequence, its terms may be denoted by \(a_n=a(n)\), and the sequence may be written \((a_n)_{n=0}^{\infty}\), \((a_n)_{n\geq 0}\), or simply \((a_n)\) when its index set is understood. The symbol \(a_n\) denotes one term; the parenthesized indexed notation denotes the whole sequence.

The distinction between a term and a sequence is similar to the distinction between a function value and the function itself. For example, \(a_4\) is one real number, whereas \((a_n)_{n=0}^{\infty}\) names all the values assigned by the sequence. The parentheses are not set braces: a sequence retains the order and the indices of its terms, while a set of values does not.

Subscripts are labels, not exponents. Thus \(a_3\) means the term at index \(3\), while \(a^3\) would ordinarily mean the cube of a quantity \(a\). Likewise, \(a_{n+1}\) is the term at the next index, whereas \(a_n+1\) is the current term plus one. Keeping the subscript grouped correctly prevents a common source of errors.

Worked Example: Reading an Indexed Formula

Define \(u_n=\frac{2}{n+1}\) for every \(n\in\mathbb{N}_0\). The index belongs in the denominator as part of \(n+1\). Substituting the indices \(0,1,2,3\) gives

$$ u_0=\frac{2}{0+1}=2,\qquad u_1=\frac{2}{1+1}=1,\qquad u_2=\frac{2}{2+1}=\frac{2}{3},\qquad u_3=\frac{2}{3+1}=\frac{1}{2}. $$

So the sequence can be displayed as \((u_n)_{n=0}^{\infty}=(2,1,\frac{2}{3},\frac{1}{2},\ldots)\). In contrast, \(u_2+1=\frac{2}{3}+1=\frac{5}{3}\), which is not \(u_3\). The term \(u_3\) is found by substituting \(3\) into the entire defining formula.

What an Ellipsis Does—and Does Not—Say

A displayed list such as \(a_0,a_1,a_2,\ldots\) indicates that further terms follow. The ellipsis is useful when a formula or a clearly established pattern tells us how the sequence continues. It is not itself a definition of the unshown terms. In particular, a few initial terms alone do not determine a unique infinite sequence.

For instance, the first four values \(5,8,11,14\) are consistent with the formula \(a_n=5+3n\), but they are also consistent with many other rules that agree at indices \(0,1,2,3\) and differ later. A sequence is fully specified only when its value is assigned at every index, either by an explicit formula, a recursive rule with suitable starting data, or another complete definition.

Worked Example: Matching a List to Its Indices

Suppose \(v_n=n^2-2n+4\) for \(n\in\mathbb{N}_0\). To write the first five terms, evaluate the rule at each of the indices \(0\) through \(4\):

$$ \begin{aligned} v_0&=0^2-2(0)+4=4,\\ v_1&=1^2-2(1)+4=3,\\ v_2&=2^2-2(2)+4=4,\\ v_3&=3^2-2(3)+4=7,\\ v_4&=4^2-2(4)+4=12. \end{aligned} $$

The indexed list is therefore \(v_0,v_1,v_2,v_3,v_4=4,3,4,7,12\), and the sequence begins \((4,3,4,7,12,\ldots)\). The repeated value \(4\) occurs at two different indices. The list is in index order; it is not a set of distinct values.

Sometimes a sequence is presented with indices starting at \(1\), rather than at \(0\). This is a valid convention when the domain is stated. But a formula must be interpreted with its stated index set: the first allowed input changes, and so does the first term. To compare two descriptions that start at different indices, match the indices explicitly instead of assuming their first displayed entries have the same label.

Worked Example: Comparing Two Starting-Index Conventions

Let \(x_n=3n-1\) for \(n\geq 1\), and let \(y_n=3n+2\) for \(n\in\mathbb{N}_0\). Their first few terms are

$$ x_1=3(1)-1=2,\quad x_2=3(2)-1=5,\quad x_3=3(3)-1=8, $$ $$ y_0=3(0)+2=2,\quad y_1=3(1)+2=5,\quad y_2=3(2)+2=8. $$

The first entries agree, but their subscripts differ. For every \(n\in\mathbb{N}_0\), substitution gives

$$ x_{n+1}=3(n+1)-1=3n+3-1=3n+2=y_n. $$

Thus the sequence indexed from \(1\) has the same terms, in the same order, as the sequence indexed from \(0\), after matching \(x_{n+1}\) with \(y_n\). The equality is between those matched terms, not between terms bearing the same subscript.

Shifted Sequences and Tails

For a fixed index \(m\), the terms from \(a_m\) onward form a tail of the original sequence. To write this tail as a new sequence whose indices begin at \(0\), shift the old indices by \(m\). This is useful when a claim concerns all terms after a particular point, rather than the entire sequence.

Definition: Let \(a:\mathbb{N}_0\to\mathbb{R}\), and let \(m\in\mathbb{N}_0\). The tail of \(a\) starting at \(m\) is the sequence \(T_m(a)\) defined by \((T_m(a))_n=a_{m+n}\) for every \(n\in\mathbb{N}_0\). In particular, its terms are \(a_m,a_{m+1},a_{m+2},\ldots\).

The shift in the formula matters: the tail’s index \(0\) corresponds to the original index \(m\), its index \(1\) corresponds to original index \(m+1\), and so on. When \(m=0\), the tail is the original sequence. Writing out the first few terms is a reliable way to check that the shift has been applied in the intended direction.

Theorem: Let \(a\) and \(b\) be real sequences, and let \(m\in\mathbb{N}_0\). Their tails starting at \(m\) are equal if and only if \(a_k=b_k\) for every integer \(k\geq m\).

Proof. First suppose \(T_m(a)=T_m(b)\). By the definition of the tails, their terms at index \(n\) are \(a_{m+n}\) and \(b_{m+n}\), respectively. Equal sequences have equal terms at every index, so \(a_{m+n}=b_{m+n}\) for every \(n\in\mathbb{N}_0\). Given any integer \(k\geq m\), the difference \(n=k-m\) is a nonnegative integer. Substituting it gives \(a_k=b_k\).

Conversely, suppose \(a_k=b_k\) for every integer \(k\geq m\). For any \(n\in\mathbb{N}_0\), \(m+n\geq m\), so \(a_{m+n}=b_{m+n}\). These are exactly the terms of \(T_m(a)\) and \(T_m(b)\) at index \(n\). The sequences have the same index set and agree at every index; by the equality theorem from “Sequences as Functions,” \(T_m(a)=T_m(b)\). This proves both directions. \(\square\)

The theorem makes precise the phrase “the sequences agree from some point onward.” Such agreement does not require the original sequences to be equal: their initial terms may differ. It says exactly that, for some \(m\), the tails starting at \(m\) are equal.

Worked Example: Finding the Matching Tails

Define \(p_n=n+1\) and \(q_n=n^2-2n+3\) for \(n\in\mathbb{N}_0\). The first terms are

$$ p_0=1,\quad p_1=2,\quad p_2=3,\quad p_3=4, $$ $$ q_0=3,\quad q_1=1-2+3=2,\quad q_2=4-4+3=3,\quad q_3=9-6+3=6. $$

The sequences differ at index \(0\), since \(p_0=1\neq 3=q_0\). For any \(n\in\mathbb{N}_0\), however,

$$ p_{n+1}=(n+1)+1=n+2, $$ $$ q_{n+1}=(n+1)^2-2(n+1)+3 =n^2+2n+1-2n-2+3=n^2+2. $$

These expressions are not equal for every \(n\), so these particular sequences do not have equal tails starting at \(1\). Checking later terms alone would not be enough to claim they eventually agree; the tail theorem requires agreement at every index in the tail. This example illustrates why the quantified condition in a tail claim cannot be replaced by checking only a few displayed terms.

Finite Initial Terms and Boundedness

A related use of tail notation is to separate a sequence into a finite initial portion and everything that follows. For boundedness, the finite initial portion cannot create an unbounded collection of values: it contains only finitely many real numbers. The following result formalizes why deleting finitely many initial terms does not change whether a sequence is bounded.

Theorem: Let \(a\) be a real sequence and \(m\in\mathbb{N}_0\). The sequence \(a\) is bounded if and only if its tail \(T_m(a)\) is bounded.

Proof. Suppose first that \(a\) is bounded. Then some \(C\geq0\) satisfies \(|a_n|\leq C\) for every \(n\in\mathbb{N}_0\). For every \(n\in\mathbb{N}_0\), the \(n\)-th term of the tail is \(a_{m+n}\), so \(|(T_m(a))_n|=|a_{m+n}|\leq C\). Thus the tail is bounded.

Conversely, suppose \(T_m(a)\) is bounded. Choose \(C\geq0\) such that \(|a_{m+n}|\leq C\) for every \(n\in\mathbb{N}_0\). If \(m=0\), this already bounds every term of \(a\). If \(m>0\), the remaining initial terms are \(a_0,\ldots,a_{m-1}\), a finite list of real numbers. Let \(D\) be the maximum of \(C\) and the absolute values of those initial terms. Then \(D\geq0\); for \(k<m\), \(|a_k|\leq D\) by its definition, and for \(k\geq m\), \(|a_k|\leq C\leq D\) by the tail bound. Therefore \(|a_k|\leq D\) for every \(k\in\mathbb{N}_0\), so \(a\) is bounded. This proves both implications. \(\square\)

This theorem concerns boundedness, not equality. Two sequences can have equal tails and different initial terms, yet one cannot be bounded while the other is unbounded: a finite set of initial terms has a finite bound. The result is useful when a proof becomes simpler after discarding a finite number of terms, provided the property being studied is one for which that deletion is justified.

A common notation pitfall is to leave the index set implicit when it matters. The expressions \((a_n)_{n=0}^{\infty}\) and \((a_n)_{n=1}^{\infty}\) assign different meanings to the starting index, even if the same formula is used. Another is to treat an ellipsis as proof that a pattern continues. A formula or definition must justify all terms, and claims about a tail require checking every index in that tail.

Check Your Understanding

Use the index labels explicitly in each answer. In particular, distinguish one term from the whole sequence.

  1. In the notation \((a_n)_{n=0}^{\infty}\), what does \(a_4\) denote, and what does the parenthesized expression denote?
  2. If \(r_n=2n^2+1\) for \(n\in\mathbb{N}_0\), calculate \(r_0,r_1,r_2\). Explain the difference between \(r_1+1\) and \(r_2\).
  3. Write the terms of \(T_2(a)\) at its indices \(0,1,2\) using terms of the original sequence \(a\).
  4. State the condition on \(a_k\) and \(b_k\) that is equivalent to \(T_3(a)=T_3(b)\).
  5. Why does deleting finitely many initial terms preserve boundedness? Identify the role played by the finite initial list.

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