Proving an Arithmetic Sequence Converges

What the Common Difference Tells Us

A constant sequence stays at one value, so its error from that value is always zero. An arithmetic sequence changes by the same amount at each step. If that amount is nonzero, the terms keep moving in one direction rather than settling near a real number. This tutorial uses the epsilon-N definition to distinguish these cases and prove exactly when an arithmetic sequence converges.

Recall that a sequence \((a_n)\) is arithmetic with common difference \(d\) when \(a_{n+1}-a_n=d\) for every \(n\in\mathbb{N}_0\). The formula \(a_n=a_0+nd\) for an arithmetic sequence was established in the earlier tutorial on arithmetic sequences. It will let us calculate the error from a proposed limit directly. We will also use two earlier results: a convergent sequence is bounded, and an arithmetic sequence with nonzero common difference is unbounded.

The formula suggests the key distinction. When \(d=0\), the term \(a_n\) does not depend on \(n\), and the constant-sequence convergence theorem applies. When \(d\neq0\), the term \(a_n=a_0+nd\) is unbounded, so it cannot converge: every convergent sequence is bounded. We will first verify the zero-difference case directly, then prove carefully why no nonzero common difference can give convergence.

The Case of Zero Common Difference

For \(d=0\), the explicit formula gives \(a_n=a_0+n\cdot0=a_0\) for every \(n\in\mathbb{N}_0\). Thus the error from \(a_0\) is zero at every index. The epsilon-N proof is short, but it still needs to specify an arbitrary tolerance, a threshold, and the required inequality on the entire tail.

Proof. Let \(\varepsilon>0\) be arbitrary. Choose \(N=0\). For every \(n\geq N\), the assumption \(d=0\) gives

This verifies the epsilon-N condition for the arbitrary positive tolerance \(\varepsilon\). Therefore \(a_n\to a_0\). \(\square\)

There is no other possible limit in this case. The sequence is constant with value \(a_0\), so the earlier theorem on the limit of a constant sequence says it converges only to \(a_0\). Equivalently, if a real number \(L\) differs from \(a_0\), then the error is the fixed positive distance \(|a_0-L|\), which cannot become smaller than every positive tolerance.

Why a Nonzero Common Difference Rules Out Convergence

For \(d\neq0\), the terms move steadily upward when \(d>0\) and steadily downward when \(d<0\), as established by the monotonicity results for arithmetic sequences. Monotonicity alone does not prove nonconvergence: a monotone sequence can converge. Here the decisive feature is that the terms are unbounded. One way to conclude nonconvergence is to cite the earlier theorem that every convergent sequence is bounded together with the earlier result that an arithmetic sequence is bounded only when its common difference is zero.

The following proof makes the failure of the epsilon-N condition explicit. It shows that, for any proposed real limit and any threshold, there is a later term whose distance from that proposed limit is at least \(1\). The Archimedean property of the real numbers ensures that a nonnegative integer can be chosen larger than any prescribed real bound.

Proof. Fix any proposed limit \(L\in\mathbb{R}\). We will show that the epsilon-N condition fails for the particular tolerance \(\varepsilon_0=1\). Let \(N\in\mathbb{N}_0\) be any proposed threshold. Since \(|d|>0\), the real number

is well-defined. By the Archimedean property, choose an integer \(n\in\mathbb{N}_0\) such that \(n>R\). Then \(n>N-1\), so \(n\geq N\), and

The triangle inequality implies the reverse estimate \(|x+y|\geq |x|-|y|\): indeed, \(|x|=|(x+y)-y|\leq|x+y|+|y|\). Apply this with \(x=nd\) and \(y=a_0-L\). Since \(a_n-L=nd+(a_0-L)\), we obtain

Thus for every proposed threshold \(N\), there is an \(n\geq N\) for which \(|a_n-L|\) is not less than \(\varepsilon_0\). No threshold can make the required inequality hold for all later indices. The epsilon-N definition therefore rules out convergence to this \(L\). Since \(L\) was arbitrary, the sequence does not converge to any real number. \(\square\)

The Convergence Criterion

The two cases give a complete classification. The zero-difference proof establishes convergence to the initial value. The nonzero-difference theorem rules out every real limit. In particular, it is not enough for an arithmetic sequence to be monotone or to have a small common difference: any fixed nonzero change persists at every step and eventually takes the terms far from any proposed limit.

Proof. If \(d=0\), the theorem for zero common difference proves that \(a_n\to a_0\). Conversely, if \(d\neq0\), the theorem on nonzero common difference shows that the sequence does not converge to any real number. Therefore convergence is possible exactly when \(d=0\), and in that case the limit is \(a_0\). \(\square\)

This conclusion also follows from the earlier results on boundedness: every convergent sequence is bounded, while an arithmetic sequence with nonzero common difference is unbounded. The explicit failure-witness proof above adds a useful detail: it identifies one fixed tolerance that defeats every proposed limit and every proposed threshold. This distinction is often useful when a question asks not merely whether a sequence fails to converge, but how the epsilon-N definition fails.

Reading the Formula Before Choosing a Threshold

A reliable proof begins by identifying the common difference and simplifying the formula for the terms. If the difference is zero, compare each term directly with \(a_0\); no estimate involving \(n\) is needed. If the difference is nonzero, do not try to find a threshold that makes the terms close to a proposed limit. Instead, fix a positive tolerance and show that every proposed threshold has a later index where the error exceeds it.

A common pitfall is to infer convergence from the fact that successive terms differ by a fixed amount. Convergence requires the terms eventually to stay within every positive tolerance of one real number. A fixed nonzero difference does not shrink as the index increases. Even when \(|d|\) is very small, the terms eventually move more than any fixed distance from a proposed limit.

Another pitfall is to prove only that the sequence is not eventually constant. That fact by itself does not rule out convergence: many convergent sequences continue to change at every step. What matters here is the unbounded linear formula, or, more explicitly, the failure-witness argument. The distinction between “the terms change” and “the terms cannot remain close to a limit” is central to a correct convergence proof.