Convergence When the Error Never Changes
In the previous tutorial, we used an estimate for \(|a_n-L|\) to choose a threshold that depends on a given tolerance. A constant sequence is a useful special case: its distance from the proposed limit does not change with \(n\). This makes the epsilon-N proof especially direct, while still requiring the same quantifiers as every other convergence proof.
Recall the epsilon-N definition: a sequence \((a_n)\) converges to \(L\) if, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}_0\) such that \(|a_n-L|<\varepsilon\) for every \(n\geq N\). For a constant sequence, the key is to compute this distance before trying to choose \(N\).
If \(a_n=A\) and the proposed limit is also \(A\), then the error is \(|a_n-A|=|A-A|=0\) at every index. Since \(0<\varepsilon\) for every positive tolerance, the required inequality holds not just eventually, but from the very first index. We can therefore choose \(N=0\) for every \(\varepsilon>0\).
Proof. Let \(\varepsilon>0\) be arbitrary. Choose \(N=0\), which belongs to \(\mathbb{N}_0\). For every \(n\geq N\), the sequence is constant, so
Thus for every positive \(\varepsilon\), there is a nonnegative integer \(N\) such that the required inequality holds for every \(n\geq N\). By the epsilon-N definition, \(a_n\to A\). \(\square\)
Worked Examples
Worked Example: A Constant Sequence with a Negative Value
Define \(a_n=-7\) for every \(n\in\mathbb{N}_0\). We claim that \(a_n\to -7\). Let \(\varepsilon>0\), and choose \(N=0\). For every \(n\geq0\),
This verifies the definition for the arbitrary tolerance \(\varepsilon\), so the sequence converges to \(-7\). The negative value causes no extra difficulty because the error is an absolute value and is exactly zero.
Worked Example: Recognizing a Constant Polynomial Expression
For \(n\in\mathbb{N}_0\), let
Expanding and cancelling gives, for every such \(n\),
Thus \((b_n)\) is the constant sequence with value \(16\). Given any \(\varepsilon>0\), take \(N=0\). Then for every \(n\geq0\),
Therefore \(b_n\to16\). The important first step was to simplify the formula exactly; the expression depends on \(n\) syntactically, but its value does not depend on \(n\).
Worked Example: A Constant Sequence Written as a Quotient
Define \(c_n=\dfrac{3n+6}{n+2}\) for \(n\in\mathbb{N}_0\). The denominator is positive because \(n+2\geq2\). Also \(3n+6=3(n+2)\), so division by the nonzero denominator gives
for every \(n\in\mathbb{N}_0\). To prove \(c_n\to3\), let \(\varepsilon>0\) and choose \(N=0\). For every \(n\geq0\),
Hence \(c_n\to3\). The domain check matters: cancellation is justified here because \(n+2\) cannot be zero for any permitted index.
Why the Threshold Does Not Depend on the Tolerance
In many convergence proofs, a smaller \(\varepsilon\) forces us to choose a later threshold. For example, the reciprocal-bound theorem in the previous tutorial produces a threshold from the size of \(\varepsilon\). Here there is no such trade-off: the error is identically zero, and zero is less than every positive tolerance. Thus the same threshold \(N=0\) works simultaneously for all \(\varepsilon>0\).
More generally, once an estimate holds for every index, it certainly holds on every tail. So any \(N\in\mathbb{N}_0\) is a valid threshold for the constant sequence converging to its value. Choosing \(N=0\) is simply the earliest and clearest choice. This is consistent with the larger-threshold principle from earlier in the course: increasing a valid threshold cannot invalidate an estimate that holds at all later indices.
The proof still has to begin with an arbitrary \(\varepsilon>0\). It is not enough to say that a constant sequence “obviously stays close” to its value. The definition asks for a threshold and an inequality verified for every index beyond it. Writing \(N=0\) and \(0<\varepsilon\) supplies both parts explicitly.
The Constant Value Is the Only Possible Limit
Convergence to the sequence’s constant value is only half the story. A constant sequence cannot converge to a different real number. This can be seen directly by choosing a tolerance that is smaller than the fixed distance between the constant value and the proposed alternative limit.
Proof. If \(L=A\), the preceding theorem gives \(a_n\to L\). Conversely, suppose \(L\neq A\). Then \(|A-L|>0\), so define the positive tolerance
For every \(n\in\mathbb{N}_0\), we have \(a_n=A\), and therefore
Consequently, for this particular positive tolerance, no index \(N\) can make \(|a_n-L|<\varepsilon_0\) hold for every \(n\geq N\): the inequality fails at every index. This contradicts the epsilon-N definition of convergence to \(L\). Therefore \(a_n\to L\) is impossible when \(L\neq A\), proving the equivalence. \(\square\)
This argument is a specific use of the Failure-Witness Criterion from earlier in the course. The fixed tolerance \(\varepsilon_0\) witnesses failure because the error never gets smaller. It also agrees with the Uniqueness of Limits theorem: the sequence converges to \(A\), so it cannot have a distinct limit. The direct proof here shows exactly which tolerance rules out a proposed wrong limit.
Worked Example: Ruling Out a Proposed Limit
The constant sequence \(d_n=5\) converges to \(5\). To test the proposed limit \(L=4\), the distance at every index is
Choose \(\varepsilon_0=\tfrac12\). Then \(|d_n-4|=1>\tfrac12\) for every \(n\in\mathbb{N}_0\). No threshold can make the error less than \(\tfrac12\) on a tail, so \(d_n\) does not converge to \(4\). The choice of tolerance is decisive: it is fixed and strictly smaller than the permanent distance from the proposed limit.
A Reliable Proof Pattern and Common Pitfalls
For a sequence given by a formula, first determine whether its terms really are constant. If they are, identify the constant value \(A\), and then compare \(a_n\) with the proposed limit. When the proposed limit is \(A\), the error calculation should end with exactly zero. There is no need to solve an inequality for \(n\), since the estimate is already true at every index.
Show that \(a_n=A\) for every permitted index, simplifying the formula with any required domain checks.
Let \(\varepsilon>0\) be arbitrary, as required by the definition of convergence.
Set \(N=0\), since the error estimate will hold from the first index.
For an arbitrary \(n\geq N\), calculate \(|a_n-A|=0<\varepsilon\).
A common mistake is to use \(|a_n-A|\leq\varepsilon\) as the final line. The definition requires a strict inequality, and equality with \(\varepsilon\) would not be enough. For the correct limit of a constant sequence, this issue is avoided because \(0<\varepsilon\) strictly for every positive \(\varepsilon\).
Another mistake is to claim convergence to any number merely because the sequence does not change. Constancy says that the terms stay at \(A\), not that they get close to every proposed \(L\). If \(L\neq A\), their error remains the positive number \(|A-L|\); choosing half that distance gives a tolerance that the sequence never meets. Together, the two directions show that a constant sequence has exactly one limit, namely its constant value.
Check Your Understanding
Use the epsilon-N definition and the fixed-error reasoning from this tutorial.
- For a sequence with \(a_n=A\) at every index, why does \(N=0\) work for every \(\varepsilon>0\) when the proposed limit is \(A\)?
- If \(b_n=(n+2)^2-n^2-4n\), what constant value does the sequence have, and what limit follows?
- Why must the denominator be checked before cancelling \(n+2\) in \(\dfrac{3n+6}{n+2}\)?
- If \(a_n=A\) and \(L\neq A\), which positive tolerance can be used to show that \(a_n\) does not converge to \(L\)?
- Why does a fixed positive error prevent an epsilon-N proof for a tolerance smaller than that error?