Writing Complete Epsilon-N Proofs

What Makes an Epsilon-N Proof Complete?

The previous tutorial focused on techniques for analyzing sequence errors. This tutorial focuses on turning those techniques into complete epsilon-N proofs. A proof of $a_n\to L$ must work for every $\varepsilon>0$: after $\varepsilon$ is given, it must specify an integer $N\in\mathbb{N}_0$ such that $|a_n-L|<\varepsilon$ for every $n\geq N$. The threshold may depend on $\varepsilon$, but it must not depend on the later index $n$.

A reliable proof has four parts: start with an arbitrary positive tolerance, select $N$, take an arbitrary $n\geq N$, and verify the required strict inequality. The order matters. An estimate that holds only at $n=N$, or a threshold chosen before $\varepsilon$ is known, does not complete the argument. Nor is it enough to say that an error “gets small”: the proof must show how the chosen threshold guarantees the required bound.

Definition: To prove $a_n\to L$ using the epsilon-N definition, let $\varepsilon>0$ be arbitrary, choose $N\in\mathbb{N}_0$ depending on $\varepsilon$, and prove that $|a_n-L|<\varepsilon$ for every $n\geq N$.

Build the Proof in Quantifier Order

The order of the definition suggests the order of the proof. First, declare $\varepsilon>0$ arbitrary. Next, inspect an estimate for $|a_n-L|$ and decide what condition on $n$ would make that estimate less than $\varepsilon$. Use that condition to choose $N$. Finally, assume $n\geq N$ and verify the estimate. The last step is essential: a proposed threshold is not justified until it is shown to work for every index in its tail.

Fix the tolerance.
Let $\varepsilon>0$ be arbitrary. Keep it visible when choosing the threshold.

Choose an integer threshold.
Use the error estimate and the Archimedean property to choose $N\in\mathbb{N}_0$ satisfying the needed inequality.

Check the whole tail.
Let $n\geq N$ be arbitrary and use $n+1\geq N+1$, or the corresponding index comparison, in the estimate.

Conclude explicitly.
Show $|a_n-L|<\varepsilon$, then state that the epsilon-N definition gives $a_n\to L$.

The theorem Choosing a Threshold from a Reciprocal Bound, established earlier in this course, gives a standard way to choose $N$ when an error is bounded by $C/(n+1)$. The examples below emphasize a broader proof-writing point: even when the threshold is found by working backward from the desired inequality, the written proof must still check the chosen integer threshold and every $n\geq N$.

Worked Example: A Rational Sequence with a Nonzero Limit

Define $a_n=(7n+1)/(3n+4)$ for $n\in\mathbb{N}_0$. We will prove $a_n\to 7/3$. Subtract the proposed limit and simplify the error. Since $3n+4>0$,

\left|a_n-\frac{7}{3}\right| = \left|\frac{3(7n+1)-7(3n+4)}{3(3n+4)}\right| = \frac{25}{3(3n+4)}.

Here the numerator simplifies because $3(7n+1)-7(3n+4)=21n+3-21n-28=-25$. Also, $3n+4\geq3(n+1)$, so

\left|a_n-\frac{7}{3}\right| \leq \frac{25}{9(n+1)} \leq \frac{3}{n+1}.

Now let $\varepsilon>0$. By the Archimedean property, choose $N\in\mathbb{N}_0$ such that $N+1>3/\varepsilon$. For any $n\geq N$, $n+1\geq N+1$, and therefore

\left|a_n-\frac{7}{3}\right| \leq \frac{3}{n+1} \leq \frac{3}{N+1} <\varepsilon.

This verifies the definition for every $n\geq N$, so $a_n\to7/3$. Notice that the choice of $N$ is made after $\varepsilon$ is given, and that the final inequality is strict even though the error estimate used non-strict inequalities.

Turn Reciprocal-Power Bounds into Thresholds

An estimate with a higher power of $n+1$ can be handled by a deliberately simple threshold. There is no need to solve for an exact or smallest $N$. Since $n+1\geq1$, raising $n+1$ to a positive integer power can only make it larger. Thus it is enough to make $N+1$ itself larger than the constant-to-tolerance ratio.

Theorem (Convergence from a Reciprocal-Power Error Bound): Let $p$ be a positive integer, let $C>0$, and suppose $$ |a_n-L|\leq\frac{C}{(n+1)^p} $$ for every $n\in\mathbb{N}_0$. Then $a_n\to L$.

Proof. Let $\varepsilon>0$ be arbitrary. By the Archimedean property, choose $N\in\mathbb{N}_0$ such that $N+1>\max\{1,C/\varepsilon\}$. Then $N+1>1$, so $(N+1)^p\geq N+1$, because $p$ is a positive integer. Also, $N+1>C/\varepsilon$, and hence

\frac{C}{(N+1)^p} \leq \frac{C}{N+1} <\varepsilon.

Now let $n\geq N$. Then $n+1\geq N+1>0$, so $(n+1)^p\geq(N+1)^p$. Using the assumed error bound gives

|a_n-L| \leq\frac{C}{(n+1)^p} \leq\frac{C}{(N+1)^p} <\varepsilon.

Thus the epsilon-N condition holds for every $n\geq N$, and $a_n\to L$. $\square$

Worked Example: A Sequence with Quadratic Denominator

Let $b_n=(n+3)/(n^2+4)$. We prove $b_n\to0$. For $n\geq1$, $n+3\leq4n$ and $n^2+4\geq n^2$, so the numerator and denominator are positive and

|b_n| =\frac{n+3}{n^2+4} \leq\frac{4n}{n^2} =\frac{4}{n}.

Given $\varepsilon>0$, choose an integer $N\geq1$ with $N>4/\varepsilon$. Such an integer exists by the Archimedean property. For every $n\geq N$, the estimate applies and $n\geq N$, so

|b_n|\leq\frac{4}{n}\leq\frac{4}{N}<\varepsilon.

Therefore $b_n\to0$. The restriction $n\geq1$ in the estimate has not been ignored: the threshold was chosen with $N\geq1$, ensuring that every index $n\geq N$ lies in the range where the estimate was proved.

This example illustrates a frequent detail in complete proofs: an estimate may only be valid after some initial index. The threshold must be chosen large enough to satisfy both requirements—the index from which the estimate applies and the index needed to make the error smaller than $\varepsilon$. If several threshold requirements arise, the earlier theorem Combining Finitely Many Thresholds allows one threshold to meet all of them.

Allocate the Tolerance Across Several Errors

A proof may involve multiple error terms rather than one reciprocal bound. If an error is a sum of $r$ parts, a direct way to manage it is to assign each part a share of the tolerance. For example, requiring each of $r$ nonnegative error bounds to be less than $\varepsilon/r$ makes their total less than $\varepsilon$. This is an explicit error budget: the shares add up to the requested tolerance.

Theorem (Finite Reciprocal Error Budget): Let $r$ be a positive integer, let $C_1,\ldots,C_r\geq0$, and suppose $$ a_n-L=\sum_{j=1}^{r}u_n^{(j)} \quad\text{and}\quad |u_n^{(j)}|\leq\frac{C_j}{n+1} $$ for every $n\in\mathbb{N}_0$ and each $j\in\{1,\ldots,r\}$. Then $a_n\to L$.

Proof. Set $S=\sum_{j=1}^{r}C_j$. If $S=0$, then each $C_j=0$, since every $C_j\geq0$. The assumed inequalities imply $|u_n^{(j)}|\leq0$, so every $u_n^{(j)}=0$. It follows that $a_n-L=0$ for every $n$, and the epsilon-N condition holds with $N=0$.

Now suppose $S>0$, and let $\varepsilon>0$ be arbitrary. Choose $N\in\mathbb{N}_0$ so that $N+1>S/\varepsilon$. For each $n\geq N$, the Triangle Inequality and the assumed bounds give

|a_n-L| =\left|\sum_{j=1}^{r}u_n^{(j)}\right| \leq\sum_{j=1}^{r}|u_n^{(j)}| \leq\sum_{j=1}^{r}\frac{C_j}{n+1} =\frac{S}{n+1} \leq\frac{S}{N+1} <\varepsilon.

The last strict inequality follows from $N+1>S/\varepsilon$. Thus the same $N$ works for every $n\geq N$, which proves $a_n\to L$. $\square$

Worked Example: Combine Two Terms with Separate Bounds

Define $c_n=4+2/(n+1)-3/(n+1)^2$. To prove $c_n\to4$, write the error as the sum of two terms. Since $n+1\geq1$,

|c_n-4| \leq\frac{2}{n+1}+\frac{3}{(n+1)^2} \leq\frac{5}{n+1}.

Let $\varepsilon>0$, and choose $N\in\mathbb{N}_0$ with $N+1>5/\varepsilon$. For every $n\geq N$,

|c_n-4| \leq\frac{5}{n+1} \leq\frac{5}{N+1} <\varepsilon.

Therefore $c_n\to4$. In this proof, it was enough to combine the terms into one reciprocal bound. In other problems, separate tolerance shares can be useful when the terms have different estimates or require different threshold conditions.

Audit the Inequalities and the Indices

A complete epsilon-N proof is short only after its dependencies are clear. Before finishing, check each of the following:

Was $\varepsilon$ arbitrary? The proof must cover every positive tolerance, not just one convenient value.
Is $N$ an allowed index? It must be a nonnegative integer, even if the algebra first suggests a real-number bound.
Does $N$ depend only on $\varepsilon$ and fixed quantities? Choosing $N$ using the later index $n$ does not establish a tail condition.
Does the estimate hold on the whole tail? Verify the index comparisons for arbitrary $n\geq N$, not just for $n=N$.
Is the final inequality strict? A bound by $\varepsilon$ is not enough when the definition requires a bound strictly less than $\varepsilon$. Choose the threshold with strict room, as in $N+1>C/\varepsilon$.

It is also important not to confuse finding a plausible threshold with proving it works. The algebra used to select $N$ may run backward from $|a_n-L|<\varepsilon$, but the verification should run forward: assume $n\geq N$, apply the valid estimates, and finish with the strict inequality. That final check is what turns a threshold idea into a proof.

Check Your Understanding

For each question, identify what must be checked to make the epsilon-N argument complete.

Why must a chosen threshold $N$ be independent of the later index $n$?
If an error estimate has been proved only for $n\geq1$, what should be ensured when choosing $N$?
Suppose $|a_n-L|\leq C/(n+1)^p$, where $C>0$ and $p$ is a positive integer. Why can it be enough to choose $N+1>\max\{1,C/\varepsilon\}$?
In a proof with $r$ error terms, why can assigning each term a tolerance of $\varepsilon/r$ control their total?
Why does the final verification need to establish a strict inequality even if some preceding error estimates use $\leq$?