Tutorials › Calculus II › Sequence Terms and Indexing

Home  /  Tutorials  /  Sequence Limits

Sequence Limits · Tutorial 499 of 998

Sequence Terms and Indexing

Learn to generate sequence terms accurately and keep the index separate from the value of the term.

Beginner 9 min read

What You'll Learn

  • Interpret a sequence as a function of an integer index
  • Identify the starting index and the value of the first term
  • Generate terms by substituting valid indices into an explicit formula
  • Distinguish an index shift from a change in the sequence values
  • Read simple sequence-limit notation without confusing the index and term

Sequences Have Indices and Values

A sequence is an ordered list of values. The notation for a sequence keeps track of both the values and the integer indices that label them. Those labels matter: the first term is not necessarily \(a_1\), and the index \(n\) is not itself the value \(a_n\).

For an explicit sequence, a formula tells us how to find a term from its index. To use the formula correctly, first check which indices are allowed. Then substitute those indices one at a time. This simple discipline prevents a common source of mistakes when listing terms and interpreting limits.

Definition: A real sequence with starting index \(n_0\) is a function \(a\) whose domain is the integers \(n\ge n_0\) and whose values are real numbers. Its value at index \(n\) is written \(a_n\). The sequence may be denoted by \((a_n)_{n=n_0}^{\infty}\). The starting index is \(n_0\); the first term's value is \(a_{n_0}\).

In many examples the starting index is \(1\), but it can be \(0\) or another integer. The formula and the stated domain work together: a formula alone does not tell you which index is the first one.

Reading the Notation Carefully

In \(a_n\), the subscript \(n\) is an index. The complete expression \(a_n\) names the term at that index. For example, if \(a_n=4n-3\) for \(n\ge1\), then \(a_1=1\), \(a_2=5\), and \(a_3=9\). The index \(2\) is not the term value: the term at index \(2\) is \(5\).

Writing the starting index explicitly is especially useful when the sequence begins at zero. In \((b_n)_{n=0}^{\infty}\), the first term is \(b_0\). It is not \(b_1\), even though \(b_1\) is also a term. The starting index tells you where to begin; the formula determines the value found there.

When calculating a term, substitute the index into every occurrence of the variable in the formula. Keep parentheses around a substituted negative index, and apply the usual order of operations. If the formula contains a denominator, confirm that it is nonzero at every index in the stated domain.

Worked Example: Listing Terms from an Index of One

Worked Example: Listing Terms from an Index of One

Let \(a_n=4n-3\) for integers \(n\ge1\). Find the first five terms, and distinguish their indices from their values.

The starting index is \(1\), so substitute \(n=1,2,3,4,5\) in order:

$$ \begin{aligned} a_1&=4(1)-3=4-3=1,\\ a_2&=4(2)-3=8-3=5,\\ a_3&=4(3)-3=12-3=9,\\ a_4&=4(4)-3=16-3=13,\\ a_5&=4(5)-3=20-3=17. \end{aligned} $$

The first five terms, in order, are \(1,5,9,13,17\). Their indices are \(1,2,3,4,5\), respectively; their values are \(1,5,9,13,17\).

Check: Substituting each listed index into \(4n-3\) gives the corresponding value above. The values increase by \(4\), consistent with the formula, since increasing \(n\) by \(1\) changes \(4n-3\) by \(4\).

When the Starting Index Is Zero

If a sequence starts at \(n=0\), evaluate the formula at zero to get the first term. Do not begin at \(n=1\) merely because that is common in other examples. This is particularly important for formulas with powers, factorials, or denominators: their values at zero can differ substantially from their values at one.

Worked Example: Finding Terms When the First Index Is Zero

Let \(b_n=\dfrac{n+2}{n+1}\) for integers \(n\ge0\). Find the first four terms.

The first permitted index is \(0\). Substitute \(n=0,1,2,3\), keeping the numerator and denominator together:

$$ \begin{aligned} b_0&=\frac{0+2}{0+1}=\frac{2}{1}=2,\\ b_1&=\frac{1+2}{1+1}=\frac{3}{2},\\ b_2&=\frac{2+2}{2+1}=\frac{4}{3},\\ b_3&=\frac{3+2}{3+1}=\frac{5}{4}. \end{aligned} $$

Thus the first four terms are \(2,\dfrac32,\dfrac43,\dfrac54\), beginning at index \(0\). Each denominator is nonzero at these indices; in fact \(n+1\ge1\) for every \(n\ge0\), so the formula is defined throughout its stated domain.

Check: Evaluating the rule at the displayed indices reproduces all four values. In particular, the first term is \(b_0=2\), not \(b_1=\dfrac32\).

Why an Index Shift Changes Labels

Sometimes it is convenient to describe a sequence using a different starting index. This changes the labels used to refer to terms, so the shift must be made consistently. The relationship is exact: if a sequence starts at \(n_0\), then its term \(k\) places after the start has index \(n_0+k\).

Index-shift rule: For a sequence \((a_n)_{n=n_0}^{\infty}\), define \(c_k=a_{n_0+k}\) for integers \(k\ge0\). Then \(c_0=a_{n_0}\), \(c_1=a_{n_0+1}\), and, in general, \(c_k\) is the same value as the original term at index \(n_0+k\). The new index \(k\) counts steps from the start; it does not change the term's value.

To justify the rule, let \(k\) be any integer with \(k\ge0\). Then \(n_0+k\ge n_0\), so \(a_{n_0+k}\) is defined in the original sequence. As \(k\) runs through \(0,1,2,\ldots\), the index \(n_0+k\) runs through \(n_0,n_0+1,n_0+2,\ldots\), in the same order. Therefore \(c_k=a_{n_0+k}\) lists exactly the original values, starting with \(a_{n_0}\), but uses labels beginning at zero. This proves the rule for every allowed \(k\).

The index shift is useful for reorganizing notation, but it is not permission to change an index in only part of an expression. When translating a formula, replace the old index everywhere and preserve the specified first value.

Worked Example: Reindexing Without Changing the Values

Let \(a_n=2n+1\) for integers \(n\ge1\). Express the same ordered values using an index \(k\ge0\), and verify the first four values in both descriptions.

The original starting index is \(n_0=1\). Set \(n=1+k\), as the index-shift rule requires. The formula becomes

$$ c_k=a_{1+k}=2(1+k)+1=2k+3,\qquad k\ge0. $$

Now calculate the first four terms from each description:

$$ \begin{aligned} a_1&=2(1)+1=3, & c_0&=2(0)+3=3,\\ a_2&=2(2)+1=5, & c_1&=2(1)+3=5,\\ a_3&=2(3)+1=7, & c_2&=2(2)+3=7,\\ a_4&=2(4)+1=9, & c_3&=2(3)+3=9. \end{aligned} $$

Both descriptions give \(3,5,7,9\), in that order. The index labels differ: \(a_1\) corresponds to \(c_0\), \(a_2\) to \(c_1\), and so on.

Check: The substitution \(n=1+k\) gives \(c_k=a_{1+k}\) for every \(k\ge0\), not just for the four values shown. The general formulas therefore preserve the entire ordered list of values.

Indices and Sequence Limits

Sequence-limit notation describes what happens to the term values as the index grows. In \(\lim_{n\to\infty}a_n=L\), the variable \(n\) is the index, \(a_n\) is the term whose value depends on that index, and \(L\) is the proposed limiting value. The statement concerns the behavior of the values \(a_n\); it does not say that the index \(n\) approaches \(L\).

For example, consider \(d_n=5+\dfrac{1}{n}\) for \(n\ge1\). Its first terms are \(d_1=6\), \(d_2=\dfrac{11}{2}\), and \(d_3=\dfrac{16}{3}\). As \(n\) grows, the added amount \(1/n\) becomes small, so the term values approach \(5\). In limit notation, this is written \(\lim_{n\to\infty}d_n=5\). Here \(n\) increases without bound, while the values \(d_n\) get closer to \(5\).

For an initial-index check, suppose the same formula is instead specified for \(n\ge0\). Then \(d_0=5+\dfrac10\) is undefined, so this formula does not define a real-valued sequence on that stated domain. The formula requires \(n\ne0\). This illustrates why the domain and formula must be checked together before listing terms or discussing a limit.

Key takeaway: In a sequence, the index tells you which term to select, and the formula gives that term's value. In a sequence limit, the index grows while the term values are examined. Never substitute the proposed limit value for the index.

Common Mistakes to Avoid

A frequent error is to assume that the first term must have index \(1\). For the sequence \(b_n=\dfrac{n+2}{n+1}\) defined for \(n\ge0\), starting at \(n=1\) would produce \(\dfrac32,\dfrac43,\ldots\) and omit the actual first term \(b_0=2\).

Another error is to confuse the index and term value. For \(a_n=4n-3\), writing \(a_2=2\) treats the index as the value. The correct calculation is \(a_2=4(2)-3=5\). And when reading \(\lim_{n\to\infty}a_n=L\), it is the values \(a_n\), not the indices \(n\), that are compared with \(L\).

Before proceeding, check three things: the stated starting index, the formula's value at that index, and whether the formula is defined for every index in its domain. These checks keep the sequence and its limit notation grounded in the correct terms.

Check Your Understanding

For each question, identify the index range before calculating or interpreting the terms.

  1. For \(u_n=3n+2\), \(n\ge1\), find \(u_1,u_2,u_3\). Which numbers are indices, and which are term values?
  2. For \(v_n=\dfrac{1}{n+2}\), \(n\ge0\), find the first three terms. What is the starting index?
  3. If \(a_n\) begins at \(n=4\), what original index corresponds to the reindexed term \(c_2=a_{4+2}\)?
  4. In \(\lim_{n\to\infty}a_n=7\), what grows without bound: \(n\), \(a_n\), or \(7\)? What value do the term values approach?
  5. Does \(w_n=\dfrac{1}{n}\) define a real-valued sequence for every \(n\ge0\)? Explain using its value at the starting index.

Previous

Next

Back to Tutorials