A Proof Strategy Begins with the Quantity to Control
In the previous tutorial, limits of rational expressions were handled by combining established limit laws and checking their hypotheses. More generally, a sequence proof becomes easier when its first step is to identify exactly what must be made small. To prove \(a_n\to L\), the quantity to control is \(|a_n-L|\). Often the main work is not applying the definition, but finding a useful expression or bound for this error.
Three habits are especially useful. First, rewrite the error algebraically so that its dependence on \(n\) is visible. Second, if the error has several parts, bound each part and combine the bounds. Third, when a direct convergence argument seems difficult, look for subsequences that reveal persistent behavior. These are proof techniques, not shortcuts around hypotheses: each estimate must hold on the indices where it is used, and each subsequence argument must specify the subsequences and their limits.
Rewrite Before Estimating
A formula can hide how close its terms are to a proposed limit. Subtracting the candidate limit may reveal cancellation, a factor that tends to zero, or a denominator that grows. For a quotient, for instance, one can subtract the proposed limit using a common denominator. The resulting expression is the actual error, and it can then be estimated.
Worked Example: Find the Error in a Quotient
Consider \(a_n=(3n+2)/(3n+5)\), with \(n\in\mathbb{N}_0\). The numerator and denominator have matching leading terms, so \(1\) is a natural candidate limit. Instead of estimating \(a_n\) directly, subtract \(1\):
The denominator is positive, and \(3n+5\geq 3(n+1)\), so
The reciprocal bound tends to zero, so the Squeeze Theorem gives \(a_n-1\to0\), and hence \(a_n\to1\) by the Limit of a Sum Theorem. Equivalently, for a direct threshold choice, given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) with \(1/(N+1)<\varepsilon\). For every \(n\geq N\), \(n+1\geq N+1\), and therefore \[ |a_n-1|\leq\frac{1}{n+1}\leq\frac{1}{N+1}<\varepsilon. \] The key step was exposing the error \(3/(3n+5)\).
A useful algebraic rewrite need not involve a quotient. If \(a_n\) is already expressed as a candidate limit plus a small term, that form gives the error immediately. If it is a difference of two nearby quantities, factoring or rationalizing may turn a difficult-looking expression into a simpler one. The point is to make the distance from the proposed limit explicit before choosing an estimate.
Combine Bounds When the Error Has Several Parts
Sometimes the error does not simplify to one term. The Triangle Inequality allows us to control it by controlling its components. If \(a_n-L=u_n+v_n\), then \[ |a_n-L|\leq |u_n|+|v_n|. \] If each component becomes small, their sum does too. This approach is especially useful when the sequence formula is already a sum, or when an algebraic rewrite separates a complicated error into manageable pieces.
Worked Example: Bound a Sum of Two Errors
Define \[ b_n=2+\frac{2}{n+1}-\frac{1}{(n+1)^2},\qquad n\in\mathbb{N}_0. \] To test the candidate limit \(2\), subtract it and use the Triangle Inequality:
The last inequality holds because \(n+1\geq1\), so \(1/(n+1)^2\leq1/(n+1)\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) such that \(3/(N+1)<\varepsilon\). For every \(n\geq N\), \[ |b_n-2|\leq\frac{3}{n+1}\leq\frac{3}{N+1}<\varepsilon. \] Thus \(b_n\to2\). The estimate is deliberately not an exact simplification of the absolute value; a sufficiently strong upper bound is enough.
When selecting a bound, it is not necessary to find the smallest possible one. It must be correct, and it must become small in a way that gives a usable threshold. The theorem Choosing a Threshold from a Reciprocal Bound, established earlier in this course, is one tool for turning bounds involving \(1/(n+1)\) into explicit choices of \(N\). When several terms must each satisfy a condition, the theorem Combining Finitely Many Thresholds allows one threshold to handle all of them.
Use Subsequences to Expose Persistent Behavior
An estimate is not always the most efficient route to a conclusion. If terms continue to display two different kinds of behavior at indices that grow without bound, it can be useful to select those indices as two subsequences. The Subsequence Theorem for convergent sequences says that every subsequence of a convergent sequence has the same limit. Together with uniqueness of limits, this gives a test for nonconvergence.
Proof. Suppose, for a contradiction, that \(a_n\to L\) for some real number \(L\). By the theorem Subsequencess of a Convergent Sequence, each of the two subsequences also converges to \(L\). But each subsequence was assumed to converge to a distinct real number. By Uniqueness of Limits, both of those numbers must equal \(L\), so they must equal each other. This contradicts their being distinct. Therefore \((a_n)\) does not converge. \(\square\)
Worked Example: Even and Odd Indices Have Different Limits
Let \(c_n=(-1)^n\) for \(n\in\mathbb{N}_0\). Taking even indices gives the subsequence \[ c_{2k}=(-1)^{2k}=1 \] for every \(k\in\mathbb{N}_0\), so this subsequence converges to \(1\). Taking odd indices gives \[ c_{2k+1}=(-1)^{2k+1}=-1 \] for every \(k\in\mathbb{N}_0\), so this subsequence converges to \(-1\). Since \(1\neq-1\), the Distinct Subsequence Limits Obstruct Convergence Theorem shows that \((c_n)\) does not converge.
This proof does not need to guess a candidate limit and then construct an epsilon witness. The two subsequences already contradict the possibility of a single limit.
For this method to work, both selected sequences of indices must define genuine subsequences: their indices increase strictly and continue indefinitely. It is also essential to verify that their limits are different. Finding subsequences that look different is not enough; the distinct-limit condition is what produces the contradiction.
A Necessary Test: Successive Differences Must Vanish
Convergence also forces neighboring terms to become arbitrarily close. This gives another technique: before attempting a full limit proof, examine \(a_{n+1}-a_n\). If these differences do not tend to zero, the original sequence cannot converge.
Proof. Suppose \(a_n\to L\). The sequence \((a_{n+1})_{n=0}^{\infty}\) is a subsequence of \((a_n)\): its indices are \(1,2,3,\ldots\), which increase strictly. By the Subsequence Theorem for convergent sequences, \(a_{n+1}\to L\). The Limit of a Difference Theorem now gives
This proves the claim. \(\square\)
Worked Example: A Successive-Difference Test for Nonconvergence
Let \(d_n=2^n\). Its successive differences are \[ d_{n+1}-d_n=2^{n+1}-2^n=2^n. \] This difference sequence does not tend to zero: it is at least \(1\) for every \(n\in\mathbb{N}_0\). If \((d_n)\) converged, the Successive Differences of a Convergent Sequence Theorem would force \(d_{n+1}-d_n\to0\). Since that necessary condition fails, \((d_n)\) does not converge.
A necessary condition is not automatically sufficient. The successive-difference test can rule out convergence when it fails, but passing the test does not establish convergence. This distinction is important: a condition that every convergent sequence must satisfy may still be shared by sequences that diverge.
Worked Example: Small Successive Differences Do Not Ensure Convergence
Consider \(e_n=\sqrt{n+1}\). Rationalizing the difference gives
The denominator is at least \(\sqrt{n+1}\), which grows without bound, so these successive differences tend to zero. Yet \((e_n)\) is unbounded: given any \(M>0\), the Archimedean property gives an integer \(n\) with \(n+1>M^2\), and then \(\sqrt{n+1}>M\). Since every convergent sequence is bounded, \((e_n)\) does not converge. Thus vanishing successive differences are necessary for convergence, but not sufficient.
Choose the Technique That Matches the Claim
These methods address different tasks. To prove convergence to a proposed \(L\), rewrite \(|a_n-L|\) and find a bound that tends to zero. To prove nonconvergence, try to identify two subsequences with distinct limits. To rule out convergence quickly, check whether successive differences fail to tend to zero. None of these techniques removes the need to check the relevant hypotheses, and a failed test for nonconvergence is not itself a proof of convergence.
A useful proof plan is therefore to name the conclusion first, then choose the evidence it requires. A convergence proof needs control of the distance to one fixed number. A distinct-subsequence argument needs two genuine subsequences and two different limits. A successive-difference argument is decisive only when the differences do not tend to zero. Making that match explicit keeps a proof focused and prevents a useful observation from being mistaken for a stronger theorem than it is.
Check Your Understanding
For each question, identify the relevant estimate or convergence principle and state what it does—and does not—allow you to conclude.
- When proving \(a_n\to L\), what quantity should an algebraic rewrite aim to simplify?
- If \(|x_n-L|\leq u_n+v_n\), with \(u_n\to0\) and \(v_n\to0\), which inequality is used to control the error?
- Why do two subsequences converging to distinct limits rule out convergence of the original sequence?
- What necessary condition does convergence impose on \(a_{n+1}-a_n\)?
- Does \(a_{n+1}-a_n\to0\) prove that \((a_n)\) converges? Give a reason for your answer.