Limit Laws for Sums

Why Limits of Sums Need an Error Estimate

The previous tutorial examined how convergence to distinct limits forces eventual ordering. We now use the same epsilon–N framework to study what happens when two sequences are combined term by term. If \(a_n\) approaches \(L\) and \(b_n\) approaches \(M\), the natural candidate for the limit of \(a_n+b_n\) is \(L+M\). The key question is how to show that the sum is close to this candidate.

The useful identity is

$$ (a_n+b_n)-(L+M)=(a_n-L)+(b_n-M). $$

It separates the total error into the error from the first sequence and the error from the second. The Triangle Inequality, established earlier in this course, then bounds the size of the total error by the sum of the two individual errors. To make that total less than a given \(\varepsilon>0\), it is enough to make each individual error less than \(\varepsilon/2\).

Proof strategy: Rewrite the error in the sum as the sum of two errors. Give each error half of the allowed tolerance, obtain a threshold for each convergence statement, and then choose an index at least as large as both thresholds.

The Limit Law for Sums

Theorem (Limit of a Sum): Suppose \(a_n\to L\) and \(b_n\to M\). Then

a_n+b_n\longrightarrow L+M.

Proof. Let \(\varepsilon>0\). Since \(a_n\to L\), the epsilon–N definition gives an \(N_1\in\mathbb{N}_0\) such that

|a_n-L|<\frac{\varepsilon}{2} \qquad\text{whenever } n\geq N_1.

Since \(b_n\to M\), there is an \(N_2\in\mathbb{N}_0\) such that

|b_n-M|<\frac{\varepsilon}{2} \qquad\text{whenever } n\geq N_2.

Choose \(N\) to be the larger of \(N_1\) and \(N_2\). For every \(n\geq N\), both estimates hold. Using the identity above and the Triangle Inequality,

\begin{aligned} |(a_n+b_n)-(L+M)| &=|(a_n-L)+(b_n-M)|\\ &\leq |a_n-L|+|b_n-M|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &=\varepsilon. \end{aligned}

This is exactly the epsilon–N condition for \(a_n+b_n\to L+M\). Therefore the termwise sum converges to the sum of the limits. \(\square\)

The proof has two separate choices to manage. The tolerance is divided so the errors add to less than \(\varepsilon\), and the two thresholds are combined so both error bounds apply to the same terms. Neither step can be omitted: an estimate for only one sequence does not control the sum, and bounds that hold after different indices must be made simultaneous.

Worked Example: Adding Two Sequences That Converge to Zero

Define \(a_n=1/(n+1)\) and \(b_n=2/(n+1)\) for \(n\in\mathbb{N}_0\). We first check their limits directly. Given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) so that \(N+1>2/\varepsilon\). For every \(n\geq N\), \(n+1\geq N+1>2/\varepsilon\), and hence

|a_n|=\frac{1}{n+1}<\frac{\varepsilon}{2}<\varepsilon, \qquad |b_n|=\frac{2}{n+1}<\varepsilon.

Thus \(a_n\to0\) and \(b_n\to0\). The Limit of a Sum theorem gives \(a_n+b_n\to0+0=0\). The formula confirms the conclusion:

a_n+b_n=\frac{1}{n+1}+\frac{2}{n+1} =\frac{3}{n+1}.

For instance, the theorem does not require us to find a new threshold by estimating this combined expression. It lets us use the already established limits of the two sequences.

Applying the Law to Nonzero Limits

The limit law does not require either limit to be zero. If the sequences approach \(L\) and \(M\), their terms are eventually close to those values, and the sum of the terms is eventually close to \(L+M\). The same half-tolerance argument works regardless of the signs or sizes of the limits.

Worked Example: A Constant and a Geometric Sequence

Let

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

The geometric-sequence convergence result established earlier in this course says that \(r^n\to0\) when \(|r|<1\). Both ratios here have absolute value \(1/3<1\). Also, constant sequences converge to their constant values, and the Limit of a Sum theorem therefore gives \(a_n\to5\) and \(b_n\to-2\). Applying that theorem once more,

a_n+b_n\longrightarrow 5+(-2)=3.

The termwise sum is

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

The powers do not cancel for every \(n\): when \(n=0\), their sum is \(1+1=2\), while when \(n=1\), it is \(1/3-1/3=0\). The limit conclusion does not depend on cancellation. Each power tends to zero, so their sum tends to zero, leaving the limit \(3\).

This example also illustrates why the theorem concerns corresponding terms: for each fixed \(n\), it adds \(a_n\) to \(b_n\). It does not pair arbitrary terms with different indices, nor does it claim that \(a_n+b_n\) equals \(L+M\) at any particular index.

Extending the Result to Finite Sums

A finite expression may contain more than two convergent sequences. The same idea applies: distribute the allowed error among the finitely many terms, then choose one threshold after which every individual estimate holds. Here is a precise version.

Theorem (Limit of a Finite Sum): Let \(r\) be a positive integer. Suppose \(a_n^{(j)}\to L_j\) for each \(j\in\{1,\ldots,r\}\). Then

\sum_{j=1}^{r}a_n^{(j)} \longrightarrow \sum_{j=1}^{r}L_j.

Proof. Let \(\varepsilon>0\). For each \(j\in\{1,\ldots,r\}\), convergence gives a threshold \(N_j\in\mathbb{N}_0\) such that

|a_n^{(j)}-L_j|<\frac{\varepsilon}{r} \qquad\text{whenever } n\geq N_j.

There are only finitely many thresholds, so one of them is largest. Let \(N=\max\{N_1,\ldots,N_r\}\). If \(n\geq N\), then \(n\geq N_j\) for every \(j\). The Triangle Inequality, applied repeatedly to the finite sum, gives

\begin{aligned} \left|\sum_{j=1}^{r}a_n^{(j)}-\sum_{j=1}^{r}L_j\right| &=\left|\sum_{j=1}^{r}\bigl(a_n^{(j)}-L_j\bigr)\right|\\ &\leq\sum_{j=1}^{r}|a_n^{(j)}-L_j|\\ &<\sum_{j=1}^{r}\frac{\varepsilon}{r}\\ &=r\frac{\varepsilon}{r}\\ &=\varepsilon. \end{aligned}

Thus the epsilon–N definition holds for the finite sum, proving the result. \(\square\)

The finiteness assumption matters in this proof. A finite collection of thresholds has a largest member, and dividing \(\varepsilon\) by the positive integer \(r\) assigns a positive error allowance to every term. An infinite collection has no general largest threshold, and an infinite sum requires additional definitions and hypotheses. This theorem makes no claim about such sums.

Worked Example: A Finite Sum with a Shared Threshold

For \(j=1,2,3,4\), define \(c_n^{(j)}=j/(n+1)\). Each sequence converges to zero. To verify this, let \(\varepsilon>0\). Choose \(N\in\mathbb{N}_0\) such that \(N+1>4/\varepsilon\). For every \(n\geq N\) and every \(j\in\{1,2,3,4\}\), we have

|c_n^{(j)}| =\frac{j}{n+1} \leq\frac{4}{N+1} <\varepsilon.

In particular, all four sequences converge to zero, so the finite-sum theorem gives

\sum_{j=1}^{4}c_n^{(j)} \longrightarrow \sum_{j=1}^{4}0=0.

The terms can also be combined explicitly:

\sum_{j=1}^{4}c_n^{(j)} =\frac{1+2+3+4}{n+1} =\frac{10}{n+1}.

For a direct check of the resulting limit, if \(n\geq N\) with \(N+1>10/\varepsilon\), then \(10/(n+1)\leq10/(N+1)<\varepsilon\). The theorem and the direct estimate agree; the theorem is especially useful when simplifying a long expression is inconvenient.

What the Sum Law Does—and Does Not—Say

The Limit of a Sum theorem is a statement about sequences of real numbers, not about adding the terms of one sequence over all indices. For example, \((a_n+b_n)\) is the sequence whose \(n\)th term is \(a_n+b_n\). By contrast, an expression such as \(a_0+a_1+a_2+\cdots\) involves infinitely many terms and is a different object. Convergence of the individual terms alone does not justify conclusions about that infinite sum.

Another common pitfall is to assume that the sum law requires the sequences to be bounded. No such assumption is needed: the hypotheses are simply that both sequences converge. Earlier in this course, it was established that convergent sequences are bounded, but the proof here uses the convergence estimates themselves rather than a separate bound on all terms.

The tolerance allocation can also be adjusted. For two sequences, any positive numbers \(\varepsilon_1,\varepsilon_2\) with \(\varepsilon_1+\varepsilon_2=\varepsilon\) would work if the first error is made less than \(\varepsilon_1\) and the second less than \(\varepsilon_2\). Choosing \(\varepsilon/2\) for both is convenient and symmetric. For \(r\) sequences, \(\varepsilon/r\) is the corresponding simple equal allocation.

Finally, convergence says that the errors become small after suitable thresholds; it does not say they are zero. The identity separating the error in the sum is exact, while the estimates control its size. Keeping those two roles distinct—algebra to identify the error, inequalities to bound it—is a reliable method for proving limit laws.

Check Your Understanding

Use the epsilon–N definition and the proved sum results to answer the following questions.

If \(a_n\to L\) and \(b_n\to M\), what identity rewrites the error \((a_n+b_n)-(L+M)\) as a sum of two errors?
Why is \(\varepsilon/2\) a useful tolerance for each of two convergent sequences?
If the two convergence estimates hold after thresholds \(N_1\) and \(N_2\), how should one choose a threshold that works for both?
Suppose three sequences converge to \(2\), \(-1\), and \(4\), respectively. What is the limit of their termwise sum?
Why does the finite-sum theorem not, by itself, establish a limit law for an infinite sum?

What You'll Learn