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Sum and Scalar-Multiple Limit Laws

Learn how convergence is preserved by adding, subtracting, and multiplying sequences by constants, and how the epsilon–N definition proves each law.

Beginner 10 min read

What You'll Learn

  • State the sum and constant-multiple limit laws for convergent sequences
  • Prove the sum law by splitting an error tolerance between two sequences
  • Handle zero and nonzero constants correctly in the scalar-multiple proof
  • Derive the difference law from addition and scalar multiplication by negative one
  • Apply the laws and verify proposed limits directly from the definition
  • Avoid errors when choosing an index bound for a scaled error

Combining Limits Without Losing Control

When two sequences each approach a fixed value, it is natural to expect their sum to approach the sum of those values. The same expectation applies to differences and to multiplying every term of a sequence by a fixed constant. The limit laws make these expectations precise.

The key point is that a limit describes what happens sufficiently far along a sequence. To prove a law, we must show that for any prescribed error tolerance, the terms can be made close enough to the proposed limit. In the sum law, this requires controlling two errors at once. In the scalar-multiple law, it requires accounting for the size of the constant.

Definition: A sequence \((a_n)\) converges to \(L\) if, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that \(n\ge N\) implies \(|a_n-L|<\varepsilon\). We write \(\lim_{n\to\infty}a_n=L\).

Here \(\varepsilon\) is any positive tolerance, and \(N\) is an index after which every term is within that tolerance of \(L\). The value of \(N\) may depend on \(\varepsilon\). We use this definition to establish the laws rather than relying only on numerical patterns.

The Sum Limit Law

Theorem: If \(\lim_{n\to\infty}a_n=A\) and \(\lim_{n\to\infty}b_n=B\), then \(\lim_{n\to\infty}(a_n+b_n)=A+B\).

The proof uses the triangle inequality, which states that \(|x+y|\le |x|+|y|\) for real numbers \(x\) and \(y\). We arrange for each sequence’s error to be less than half the required total error. The two errors then add to less than \(\varepsilon\).

1
Set the desired tolerance.
Let \(\varepsilon>0\) be arbitrary. Since \(a_n\to A\), choose \(N_a\) so that \(n\ge N_a\) implies \(|a_n-A|<\varepsilon/2\). Since \(b_n\to B\), choose \(N_b\) so that \(n\ge N_b\) implies \(|b_n-B|<\varepsilon/2\).
2
Use one index for both estimates.
Set \(N=\max\{N_a,N_b\}\). If \(n\ge N\), then both estimates hold.
3
Bound the sum’s error.
By the triangle inequality, \(|(a_n+b_n)-(A+B)|\le |a_n-A|+|b_n-B|<\varepsilon/2+\varepsilon/2=\varepsilon\). This proves the sum law.

This argument explains why the limit of the sum is \(A+B\): subtracting the proposed limit separates the total error into the two individual errors. The choice \(N=\max\{N_a,N_b\}\) matters because both sequences must satisfy their bounds at the same index.

Worked Example: Adding Two Convergent Sequences

Worked Example: Adding Two Convergent Sequences

Let \(a_n=4+\frac{3}{n}\) and \(b_n=-1+\frac{2}{n}\), for \(n\in\mathbb{N}\). Find the limit of \(a_n+b_n\), and verify it directly.

First, the terms simplify exactly:

$$ a_n+b_n =\left(4+\frac{3}{n}\right)+\left(-1+\frac{2}{n}\right) =3+\frac{5}{n}. $$

Since \(1/n\to0\), the two component limits are \(a_n\to4\) and \(b_n\to-1\). The sum law therefore gives \(a_n+b_n\to4+(-1)=3\).

For a direct check, let \(\varepsilon>0\). Choose \(N=\left\lfloor 5/\varepsilon\right\rfloor+1\), where \(\lfloor x\rfloor\) is the greatest integer not exceeding \(x\). Then \(N>5/\varepsilon\). For every \(n\ge N\),

$$ |(a_n+b_n)-3| =\left|\frac{5}{n}\right| =\frac{5}{n} \le \frac{5}{N} <\varepsilon. $$

Thus the definition confirms the limit \(3\).

The Constant-Multiple Limit Law

Theorem: If \(\lim_{n\to\infty}a_n=A\) and \(c\in\mathbb{R}\) is a fixed constant, then \(\lim_{n\to\infty}(ca_n)=cA\).

The constant must be fixed: it does not vary with \(n\). If \(c\ne0\), an error in \(a_n\) is multiplied in size by \(|c|\). Therefore, to make the scaled error less than \(\varepsilon\), it is enough to make the original error less than \(\varepsilon/|c|\). If \(c=0\), every term of the scaled sequence is exactly zero, so that case is immediate.

1
Handle the zero constant.
If \(c=0\), then \(ca_n=0=cA\) for every \(n\), so the sequence converges to \(cA=0\).
2
Choose a suitable tolerance when \(c\ne0\).
Let \(\varepsilon>0\). Because \(a_n\to A\), there is an index \(N\) such that \(n\ge N\) implies \(|a_n-A|<\varepsilon/|c|\).
3
Scale the error.
For \(n\ge N\), \(|ca_n-cA|=|c|\,|a_n-A|<|c|(\varepsilon/|c|)=\varepsilon\). This proves the law for every nonzero \(c\), and the zero case was already established.

This proof works for negative constants as well as positive ones because absolute value measures the size of the factor: \(|ca_n-cA|=|c|\,|a_n-A|\). Multiplication by a negative constant reverses signs but does not change the error calculation.

Worked Example: Multiplying by a Negative Constant

Worked Example: Multiplying by a Negative Constant

Let \(a_n=2-\frac{1}{n}\). Find the limit of \(-3a_n\), and verify the answer using the definition.

First compute the scaled terms and the scaled candidate limit:

$$ -3a_n=-3\left(2-\frac{1}{n}\right)=-6+\frac{3}{n}, \qquad -3\cdot 2=-6. $$

Because \(1/n\to0\), \(a_n\to2\). The scalar-multiple law predicts \(-3a_n\to-6\). To check, let \(\varepsilon>0\) and choose \(N=\left\lfloor 3/\varepsilon\right\rfloor+1\). For \(n\ge N\), \(n>3/\varepsilon\), and hence

$$ |(-3a_n)-(-6)| =\left|-6+\frac{3}{n}+6\right| =\frac{3}{n} \le\frac{3}{N} <\varepsilon. $$

The direct estimate verifies the limit \(-6\).

The Difference Limit Law

Subtraction is covered by the two laws already proved. Since \(b_n\to B\), the constant-multiple law with \(c=-1\) gives \(-b_n\to-B\). Applying the sum law to \(a_n+(-b_n)\) gives the difference law.

Corollary: If \(a_n\to A\) and \(b_n\to B\), then \(a_n-b_n\to A-B\).

Indeed, \(a_n-b_n=a_n+(-1)b_n\), and the limits combine as \(A+(-1)B=A-B\). The law does not require either limit to be positive, or either sequence to have terms of one sign. It requires only that both sequences converge to finite real numbers.

Worked Example: Subtracting Convergent Sequences

Worked Example: Subtracting Convergent Sequences

Let \(a_n=1+\frac{1}{n}\) and \(b_n=4-\frac{2}{n}\). Find the limit of \(a_n-b_n\), then check it directly.

Subtract the expressions, keeping the signs of every term:

$$ a_n-b_n =\left(1+\frac{1}{n}\right)-\left(4-\frac{2}{n}\right) =1+\frac{1}{n}-4+\frac{2}{n} =-3+\frac{3}{n}. $$

The component limits are \(a_n\to1\) and \(b_n\to4\), so the difference law predicts \(a_n-b_n\to1-4=-3\). For a definition-based check, let \(\varepsilon>0\) and set \(N=\left\lfloor 3/\varepsilon\right\rfloor+1\). Whenever \(n\ge N\),

$$ |(a_n-b_n)-(-3)| =\left|-3+\frac{3}{n}+3\right| =\frac{3}{n} \le\frac{3}{N} <\varepsilon. $$

Therefore the difference converges to \(-3\), as claimed.

A Common Pitfall: Forgetting the Scale Factor

When multiplying a sequence by a constant, it is not enough to use the same error bound for the original sequence. The constant changes the size of the error. For example, let \(a_n=1+\frac{1}{n}\) and consider \(5a_n\). A mistaken calculation might say that \(|a_n-1|<\varepsilon\) guarantees \(|5a_n-5|<\varepsilon\). The actual relation is

$$ |5a_n-5|=5|a_n-1|. $$

For a concrete failure, take \(\varepsilon=0.1\) and \(n=20\). Then \(|a_{20}-1|=1/20=0.05<0.1\), but \(|5a_{20}-5|=5/20=0.25\), which is not less than \(0.1\). The correction is to require \(|a_n-1|<\varepsilon/5=0.02\). For instance, choosing \(N=51\) ensures that \(n\ge N\) gives \(5/n\le5/51<0.1\). In general, use \(\varepsilon/|c|\) when \(c\ne0\), not \(\varepsilon\) unchanged.

Why These Laws Matter

The laws let us combine known limits in expressions built from sums, differences, and fixed scalar multiples. They also make clear which information is needed: the component sequences must converge, and the constants must not vary with the index. This tutorial establishes only these operations; more complicated combinations require additional limit laws.

For calculations, it is useful to keep two levels distinct. The laws predict a limit from component limits. A direct epsilon–N check then measures the distance of the actual terms from that proposed limit. When checking algebra, preserve parentheses in a difference and use absolute values when measuring errors. Those habits prevent sign mistakes and lost factors.

Takeaway: If \(a_n\to A\), \(b_n\to B\), and \(c\) is a fixed real number, then \(a_n+b_n\to A+B\), \(a_n-b_n\to A-B\), and \(ca_n\to cA\). The proofs work by controlling each error and, for scalar multiplication, adjusting the tolerance by the magnitude of the constant.

Check Your Understanding

Use the definitions and limit laws to answer the following questions.

  1. If \(a_n\to 6\) and \(b_n\to -2\), what limit does the sum law give for \(a_n+b_n\)?
  2. If \(a_n\to -3\), what is the limit of \(4a_n\), and what tolerance should be imposed on \(|a_n-(-3)|\) to make the scaled error less than \(\varepsilon\)?
  3. Explain why the difference law follows from the sum and constant-multiple laws.
  4. For \(c=0\), why does the scalar-multiple proof not need to divide \(\varepsilon\) by \(|c|\)?
  5. If \(a_n=2+\frac{1}{n}\) and \(b_n=5-\frac{3}{n}\), calculate \(a_n-b_n\) and state its limit.

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