Epsilon-N Proofs for Rational Sequences

Why Prove a Sequence Limit Directly?

A calculation or a pattern in the terms may suggest the limit of a sequence, but it does not show that the terms eventually lie within every prescribed distance of that limit. An epsilon-N proof supplies that guarantee: it starts with any positive tolerance \(\varepsilon\), then gives an index \(N\) after which every term is within \(\varepsilon\) of the proposed limit.

For a rational expression in the index, the key work is often algebraic. Subtract the proposed limit, simplify the resulting fraction, and bound its numerator from above and its denominator away from zero. The resulting estimate should make the error small when \(n\) is sufficiently large.

The index threshold may depend on \(\varepsilon\), and an explicit choice of \(N\) makes the proof complete. It is not necessary to find the smallest possible threshold; it is necessary to show that the one chosen works.

A Basic Strategy for Rational Expressions

Consider a sequence \(a_n=P(n)/Q(n)\), where \(P\) and \(Q\) are polynomials and \(Q(n)\ne0\) for the indices under consideration. For a proposed limit \(L\), begin with \(|a_n-L|\). Combining the terms over a common denominator often produces a numerator of lower degree than \(Q\). Then use simple power bounds: for \(n\ge1\), \(n^k\le n^m\) whenever \(0\le k\le m\).

For example, if a denominator has leading term \(b n^q\), with \(b\ne0\), and its remaining terms have degree at most \(q-1\), those lower-degree terms can be bounded by a constant times \(n^{q-1}\). For large enough \(n\), the leading term therefore controls the size of the whole denominator. This is what makes the fraction's error small.

Worked Example: A Quotient of Linear Polynomials

Worked Example: Equal-Degree Polynomials

Worked Example: A Quotient with Limit Zero

Why the Leading Terms Determine These Limits

The examples reflect a general fact for rational expressions whose numerator's degree is no greater than the denominator's. Here the degree of a nonzero polynomial is the largest exponent with a nonzero coefficient. The leading coefficient is the coefficient of that term.

Here is an explicit bound that justifies the result. Write \(Q(n)=b_qn^q+\sum_{k=0}^{q-1}b_kn^k\), and let \(S=\sum_{k=0}^{q-1}|b_k|\). Let \(L\) be the proposed limit: \(a_q/b_q\) for equal degrees, or \(0\) for a numerator of lower degree. The polynomial \(H(n)=P(n)-LQ(n)\) has degree at most \(q-1\). Write its coefficients as \(h_k\), and set \(C=\sum_{k=0}^{q-1}|h_k|\), using \(C=0\) if \(H\) is identically zero.

For \(n\ge1\), each \(n^k\) with \(k\le q-1\) is at most \(n^{q-1}\). Applying the triangle inequality to the lower-degree terms gives \(|Q(n)-b_qn^q|\le S n^{q-1}\) and \(|H(n)|\le C n^{q-1}\). In particular, whenever \(n\ge 2S/|b_q|\),

This lower bound also proves that \(Q(n)\ne0\) for these indices. Since \(P(n)/Q(n)-L=H(n)/Q(n)\), the error is bounded as follows:

For any \(\varepsilon>0\), one explicit choice is

When \(n\ge N\), the denominator bound holds and \(n>2C/(|b_q|\varepsilon)\). If \(C>0\), the error bound is therefore strictly less than \(\varepsilon\); if \(C=0\), the error is identically zero for these indices. This establishes the epsilon-N condition. If the denominator is a nonzero constant (\(q=0\)), the only numerator degree allowed by the theorem is also zero, and the quotient is constant, so its limit is immediate.

A Common Denominator-Bound Mistake

A frequent error is to replace a denominator by its leading term as though all its lower-order terms increase its size. For example, consider \(a_n=n^2/(n^2-n)\) for \(n\ge2\). Its proposed limit is \(1\), and the exact error is

The incorrect calculation \(n^2-n\ge n^2\) would give the false bound \(|a_n-1|\le1/n\). In fact \(n^2-n<n^2\), so that denominator comparison goes in the wrong direction. For \(n\ge2\), however, \(n^2-n\ge n^2/2\), because \(n\le n^2/2\). The valid estimate is

Thus \(N=\max\{2,\lfloor2/\varepsilon\rfloor+1\}\) works: if \(n\ge N\), then \(2/n<\varepsilon\). The correction is to establish the denominator's lower bound with the lower-order terms included, rather than simply discarding them.

Putting the Proof Together

For a new rational-form sequence, organize the argument so every estimate contributes to the epsilon-N conclusion.

These proofs are useful because they show exactly how large the index must be for any requested accuracy. They also make clear which algebraic facts support the limit, rather than relying only on the appearance of the first few terms or on a formal comparison of degrees.