Choosing N From Epsilon

From a Tolerance to a Threshold

The previous tutorial focused on reading the quantifiers in the epsilon-N definition. Here the task is to choose \(N\): after a tolerance \(\varepsilon>0\) is given, we must find a nonnegative integer threshold that makes the required estimate hold for every later index. The key is to begin with an estimate for \(|a_n-L|\), then solve that estimate for \(n\).

A threshold is not guessed independently of the tolerance. If the error is bounded by a reciprocal expression, a smaller tolerance generally calls for a larger threshold. Once a candidate \(N\) is found, the proof must still verify the inequality for an arbitrary \(n\geq N\). Checking only the first such index is not sufficient unless the estimate also shows that all later indices are controlled.

The strict inequality matters. If an estimate only gives \(|a_n-L|\leq\varepsilon\), it does not establish the condition \(|a_n-L|<\varepsilon\). A threshold should therefore be chosen so the bound itself is strictly less than the tolerance. In practice, this often means choosing an integer strictly larger than a real-number expression.

A Reciprocal Error Bound

The simplest common case is an error bounded by a constant divided by \(n+1\). The following result turns that estimate directly into a threshold. The integer part \(\lfloor x\rfloor\) denotes the greatest integer less than or equal to \(x\).

Proof. Fix \(\varepsilon>0\), and set \(N=\left\lfloor C/\varepsilon\right\rfloor\). Since \(C/\varepsilon>0\), \(N\) is a nonnegative integer. By the defining property of the floor, \[ N\leq\frac{C}{\varepsilon}<N+1. \] Now take any \(n\in\mathbb{N}_0\) with \(n\geq N\). Then \(n+1\geq N+1>C/\varepsilon\). Since all quantities are positive, taking reciprocals reverses the strict inequality, so \[ \frac{C}{n+1}<\varepsilon. \] The assumed error bound now gives \[ |a_n-L|\leq\frac{C}{n+1}<\varepsilon. \] This holds for every \(n\geq N\), as required. \(\square\)

The floor in this choice is deliberate. It may appear that one should round \(C/\varepsilon\) up, but the threshold is an index from which \(n+1\) is used in the denominator. Even when \(C/\varepsilon\) is an integer, \(N=\lfloor C/\varepsilon\rfloor\) gives \(n+1\geq N+1>C/\varepsilon\), which supplies the strict inequality.

Solving Other Error Estimates

Not every useful estimate has exactly the form \(C/(n+1)\). The same strategy applies: rearrange the error bound until it becomes a condition on \(n\), then choose an integer that satisfies it. When the rearranged condition is strict, it is often safest to choose the floor of its right-hand side plus one. This guarantees an integer strictly above that real number, whether or not the real number is itself an integer.

For a squared denominator, solving the inequality introduces a square root. Because \(n\) is nonnegative, a condition on \(n^2\) can be translated into a condition on \(n\), and then an integer strictly above the resulting real number can be chosen.

Combining Several Tail Estimates

An error may be a sum of terms, each controlled by a different estimate. In that case, first allocate part of the tolerance to each term. Then choose a threshold for each part and use the largest threshold. The next result justifies this common step.

Proof. Each \(N_j\leq N\). Take any \(n\in\mathbb{N}_0\) with \(n\geq N\). Then \(n\geq N_j\) for every \(j\), so the assumed property of each \(N_j\) gives \(P_j(n)\). Since \(n\) was arbitrary, every condition holds for every \(n\geq N\). \(\square\)

This result uses the same tail logic as the theorem on larger thresholds from the previous tutorial: moving a threshold later preserves an estimate already valid on a tail. Taking the maximum gives one threshold that is late enough for every component. If the component estimates add to the total error, the triangle inequality then combines their bounds.

Choosing a Useful Bound

A sequence can have a complicated formula while its distance from the proposed limit has a simple upper bound. It is usually better to work with a convenient bound than to solve an unnecessarily complicated inequality exactly. For example, if an expression is known to be at most \(C/(n+1)\), the reciprocal-bound theorem supplies a threshold immediately, even if the original expression has extra terms in its denominator.

The bound must be valid on the indices where it is used. If an estimate is established only for \(n\geq1\), a proposed threshold of \(N=0\) would require separate attention at \(n=0\). One can instead choose \(N\geq1\), or verify the initial case as part of the argument. Similarly, denominators must be positive before multiplying or taking reciprocals, and square-root rearrangements require attention to the nonnegative range of the index.

A practical proof can be organized in four steps:

A useful threshold need not be optimal. A slightly larger integer may make the proof shorter and less vulnerable to an endpoint error. The larger-threshold theorem ensures that once an estimate holds from some \(N\), any larger threshold also works for the same tolerance. What matters is that the chosen integer controls every index in the tail and that the final inequality is strict.