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Limit Laws for Differences

Use the limit law for differences and related results to reason efficiently about subtracted sequences and their limits.

Intermediate 9 min read

What You'll Learn

  • Apply the established limit law for differences to sequences with nonzero limits
  • Determine an unknown sequence limit from a convergent sequence and its difference
  • Use uniqueness of limits when checking a proposed limit
  • Combine successive limiting differences along a finite chain
  • Distinguish termwise differences from infinite series

Subtraction as a Limit Operation

For sequences \(a_n\) and \(b_n\), their termwise difference is the sequence whose \(n\)th term is \(a_n-b_n\). If \(a_n\) approaches \(L\) and \(b_n\) approaches \(M\), the natural candidate for the difference limit is \(L-M\). Earlier in this course, the result was established under the name Limit of a Difference: if \(a_n\to L\) and \(b_n\to M\), then \(a_n-b_n\to L-M\). We will use that result rather than repeat its proof.

The law applies to arbitrary real limits, including negative limits and limits that are equal. It is often useful not only to calculate the limit of a difference, but also to recover one sequence's limit from the other sequence and their difference. We will develop that reverse use, then see how successive differences combine.

Recall (Limit of a Difference): If \(a_n\to L\) and \(b_n\to M\), then
$$ a_n-b_n\longrightarrow L-M. $$

The important point is that the same index appears in both terms: the sequence under consideration has terms \(a_n-b_n\), not differences formed by pairing terms at unrelated indices.

Worked Examples with the Difference Law

Worked Example: Subtracting Sequences with Nonzero Limits

Define

$$ a_n=7+\left(\frac{2}{5}\right)^n, \qquad b_n=1-\left(-\frac{2}{3}\right)^n. $$

Both geometric ratios have absolute value less than \(1\), so the earlier theorem on geometric sequences with a contracting ratio gives \((2/5)^n\to0\) and \((-2/3)^n\to0\). By the Limit of a Sum, the Limit of a Difference, and the convergence of constant sequences, \(a_n\to7\) and \(b_n\to1\). The Limit of a Difference therefore gives

$$ a_n-b_n\longrightarrow 7-1=6. $$

The expression can also be checked directly:

$$ a_n-b_n =7+\left(\frac{2}{5}\right)^n-\left(1-\left(-\frac{2}{3}\right)^n\right) =6+\left(\frac{2}{5}\right)^n+\left(-\frac{2}{3}\right)^n. $$

The two power terms both tend to zero, consistent with the limit \(6\). They do not cancel at every index: at \(n=0\), their sum is \(1+1=2\), while at \(n=1\), it is \(2/5-2/3=-4/15\). The limit law does not depend on cancellation at particular indices.

Worked Example: Equal Limits Give a Difference Limit of Zero

Let \(c_n=4+(1/4)^n\) and \(d_n=4-(1/3)^n\). Since both geometric power terms tend to zero, the Limit of a Sum and the Limit of a Difference give \(c_n\to4\) and \(d_n\to4\). Consequently, the Limit of a Difference gives

$$ c_n-d_n\longrightarrow 4-4=0. $$

For each \(n\), the difference is

$$ c_n-d_n =4+\left(\frac{1}{4}\right)^n -\left(4-\left(\frac{1}{3}\right)^n\right) =\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n. $$

This example illustrates a useful interpretation: when two sequences approach the same number, their termwise difference approaches zero. Their terms need not be equal, even eventually; convergence only requires that the difference become arbitrarily small.

Recovering a Limit from a Difference

Subtraction can be rearranged to solve for either of its terms. If \(d_n=a_n-b_n\), then \(b_n=a_n-d_n\) and \(a_n=b_n+d_n\). Thus, if we know limits for one sequence and for the difference, the sum and difference laws determine the limit of the remaining sequence. Uniqueness of limits ensures that a limit obtained this way is the only possible one.

Theorem (Two Limits Determine the Third): Let \(d_n=a_n-b_n\). If any two of the sequences \(a_n\), \(b_n\), and \(d_n\) converge, then the third converges as well. More precisely:
  • If \(a_n\to L\) and \(b_n\to M\), then \(d_n\to L-M\).
  • If \(a_n\to L\) and \(d_n\to D\), then \(b_n\to L-D\).
  • If \(b_n\to M\) and \(d_n\to D\), then \(a_n\to M+D\).

Proof. The first claim is exactly the Limit of a Difference recalled above. For the second, the defining relation \(d_n=a_n-b_n\) gives \(b_n=a_n-d_n\) for every \(n\). Since \(a_n\to L\) and \(d_n\to D\), the Limit of a Difference gives

$$ b_n=a_n-d_n\longrightarrow L-D. $$

For the third claim, rearrange the same relation as \(a_n=b_n+d_n\). Since \(b_n\to M\) and \(d_n\to D\), the Limit of a Sum gives

$$ a_n=b_n+d_n\longrightarrow M+D. $$

All three cases are proved. \(\square\)

Worked Example: Finding a Limit from a Known Difference

Suppose \(a_n=5+(-1/2)^n\) and define \(d_n=2+1/(n+1)\). The geometric-sequence result gives \(a_n\to5\). Also, \(1/(n+1)\to0\): given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) with \(N+1>1/\varepsilon\); then for \(n\geq N\),

$$ \left|\frac{1}{n+1}\right| \leq\frac{1}{N+1} <\varepsilon. $$

The Limit of a Sum theorem now gives \(d_n\to2\). Define \(b_n=a_n-d_n\), so that \(d_n=a_n-b_n\). The two-limits theorem yields \(b_n\to5-2=3\). Substitution confirms the terms and their limiting components:

$$ b_n=3+\left(-\frac{1}{2}\right)^n-\frac{1}{n+1}. $$

Both terms after \(3\) tend to zero, in agreement with the result. Notice that the theorem determines the limit without requiring a separate estimate for the entire expression for \(b_n\).

Combining Successive Differences

A difference can also be broken into a finite chain. For example, \(x_n-z_n\) can be written as \((x_n-y_n)+(y_n-z_n)\). The intermediate terms cancel exactly. If each successive difference has a limit, the Limit of a Finite Sum theorem then determines the limit from the sum of those individual limits. The following result states this for any finite number of steps.

Theorem (Limit Along a Finite Chain of Differences): Let \(r\) be a positive integer, and let \(x_n^{(0)},x_n^{(1)},\ldots,x_n^{(r)}\) be real sequences. Suppose, for each \(j\in\{1,\ldots,r\}\), that
$$ x_n^{(j-1)}-x_n^{(j)}\longrightarrow D_j. $$

Then

$$ x_n^{(0)}-x_n^{(r)} \longrightarrow \sum_{j=1}^{r}D_j. $$

Proof. For every \(n\), adding the successive differences cancels each intermediate term once with a positive sign and once with a negative sign:

$$ \sum_{j=1}^{r}\left(x_n^{(j-1)}-x_n^{(j)}\right) =x_n^{(0)}-x_n^{(r)}. $$

For instance, the terms \(x_n^{(1)},\ldots,x_n^{(r-1)}\) each occur in adjacent summands as \(-x_n^{(j)}+x_n^{(j)}=0\); only \(x_n^{(0)}\) and \(-x_n^{(r)}\) remain. By hypothesis, each sequence \(x_n^{(j-1)}-x_n^{(j)}\) converges to \(D_j\). The Limit of a Finite Sum theorem therefore gives

$$ \sum_{j=1}^{r}\left(x_n^{(j-1)}-x_n^{(j)}\right) \longrightarrow \sum_{j=1}^{r}D_j. $$

Using the identity above proves the claimed limit. \(\square\)

Worked Example: Adding Limits Along a Three-Step Chain

Consider the sequences

$$ x_n^{(0)}=6+\left(\frac{1}{2}\right)^n,\quad x_n^{(1)}=3+\left(\frac{1}{3}\right)^n,\quad x_n^{(2)}=1-\left(\frac{1}{4}\right)^n,\quad x_n^{(3)}=-2+\left(\frac{1}{5}\right)^n. $$

Each geometric term tends to zero. Applying the Limit of a Difference to each neighboring pair shows that the three successive differences converge to the differences of the corresponding constant limits:

$$ x_n^{(0)}-x_n^{(1)}\to 6-3=3,\qquad x_n^{(1)}-x_n^{(2)}\to 3-1=2,\qquad x_n^{(2)}-x_n^{(3)}\to 1-(-2)=3. $$

The finite-chain theorem gives \(x_n^{(0)}-x_n^{(3)}\to3+2+3=8\). Direct subtraction confirms the result:

$$ x_n^{(0)}-x_n^{(3)} =8+\left(\frac{1}{2}\right)^n-\left(\frac{1}{5}\right)^n \longrightarrow 8. $$

The chain method is useful when a complicated difference is naturally compared through intermediate sequences. It permits each smaller limiting difference to be established separately, then combines them using a finite sum.

Scope and Common Pitfalls

These results concern termwise differences of sequences. They do not, by themselves, address an infinite expression such as \(a_0-a_1+a_2-a_3+\cdots\); that is an infinite series, a different object requiring its own definitions and hypotheses. Likewise, the finite-chain theorem assumes a positive integer number of steps. Its proof uses a finite sum and makes no claim about an infinite chain.

A second point to keep clear is the order of subtraction. In general, \(L-M\) is not \(M-L\). If the order of the sequences is reversed, the limit changes sign: \(b_n-a_n\to M-L\). This follows either by applying the Limit of a Difference in the reversed order or from \(b_n-a_n=-(a_n-b_n)\); the displayed conclusion is consistent because \(M-L=-(L-M)\).

Finally, the two-limits theorem is a way to infer convergence, not permission to assume an unknown limit. Its hypotheses require that the two specified sequences really converge. Once they do, algebraic rearrangement identifies the third sequence, and an established limit law supplies its limit. When a proposed value is obtained in more than one way, the Uniqueness of Limits theorem guarantees that the answers must agree.

Check Your Understanding

Use the limit laws and the results proved here to answer the following questions.

  1. If \(a_n\to-2\) and \(b_n\to5\), what is the limit of \(a_n-b_n\)?
  2. If \(a_n\to8\) and \(d_n=a_n-b_n\to3\), what is the limit of \(b_n\), and which algebraic rearrangement gives it?
  3. Why does knowing the limits of \(b_n\) and \(d_n=a_n-b_n\) determine the limit of \(a_n\)?
  4. If three successive differences converge to \(2\), \(-1\), and \(4\), what is the limit of the difference between the first and last sequences in the chain?
  5. Why does the finite-chain theorem not establish a result for an infinite chain of differences?

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