From an Error Estimate to an Index Bound
A direct proof of a sequence limit begins with a precise goal: make every sufficiently late term lie within a prescribed distance of the proposed limit. The number that measures this allowed distance is \(\varepsilon\). The proof must work no matter how small the positive value of \(\varepsilon\) is.
For simple sequences, the central task is often to turn an estimate for the error into a condition on the index \(n\). Solving that condition suggests how large an index must be. The final step is to choose a natural number that ensures the condition holds for every later index.
The quantifiers describe the order of the proof. First, an arbitrary positive \(\varepsilon\) is given. The proof then chooses \(N\), which may depend on \(\varepsilon\). Once \(N\) has been chosen, the estimate must hold for all \(n\geq N\), not just for one index.
The Basic Strategy
Suppose the distance from \(a_n\) to a proposed limit \(L\) has a simple expression. Start with the inequality that would guarantee convergence:
Then solve this inequality for \(n\), taking care that any multiplication or division preserves the inequality direction. The result may be a real-number threshold rather than a natural number. Choose a natural number at or beyond that threshold, then check that every \(n\) at or beyond the chosen index satisfies the original error inequality.
For example, if the error has the form \(c/n\), where \(c>0\), then
These equivalences use \(n>0\) and \(\varepsilon>0\), so multiplying or dividing by either quantity does not reverse the inequality. The last condition explains what the index must do. Because \(N\) must be a natural number and the inequality is strict, a convenient choice is
where \(\lfloor x\rfloor\) is the greatest integer less than or equal to \(x\). In particular, \(\lfloor x\rfloor+1>x\). This choice is sufficient; the proof does not need to find the smallest possible index.
Worked Example: Proving \(1/n\) Tends to Zero
Worked Example: Proving \(1/n\) Tends to Zero
Let \(a_n=1/n\), with \(n\in\mathbb{N}\). We prove directly that \(a_n\) converges to \(0\).
Let \(\varepsilon>0\) be arbitrary. The required error estimate is
Since \(n>0\), solving this inequality for \(n\) gives \(n>1/\varepsilon\). Choose
This is a natural number, and the defining property of the floor gives \(N>1/\varepsilon\). If \(n\geq N\), then \(n\geq N>1/\varepsilon\). Taking reciprocals of positive quantities gives \(1/n<\varepsilon\). Therefore \(|a_n-0|<\varepsilon\) for every \(n\geq N\).
The chosen index thus meets the definition for the arbitrary \(\varepsilon>0\). Hence the sequence converges to \(0\). As a check, substituting the bound \(n\geq N>1/\varepsilon\) into the error gives \(1/n\leq 1/N<\varepsilon\), as required.
Worked Example: A Constant Plus a Reciprocal Error
Worked Example: Proving \(4+3/(n+1)\) Tends to \(4\)
Define \(a_n=4+3/(n+1)\). The difference from the proposed limit \(4\) is
Let \(\varepsilon>0\). We want \(3/(n+1)<\varepsilon\). Since \(n+1\) and \(\varepsilon\) are positive, this is equivalent to
It is enough to choose an integer \(N\) with \(N+1>3/\varepsilon\). The following choice guarantees that:
Indeed, \(N>3/\varepsilon\), so \(N+1>3/\varepsilon\) as well. For every \(n\geq N\), we have \(n+1\geq N+1>3/\varepsilon\). Taking reciprocals and multiplying by \(3>0\) gives
Thus \(|a_n-4|<\varepsilon\) for every \(n\geq N\), proving that \(a_n\) converges to \(4\). Substituting the index bound back into the error confirms the result: \(n+1\geq N+1>3/\varepsilon\), hence \(3/(n+1)<\varepsilon\).
Worked Example: Choosing a Bound for a Squared Index
Worked Example: Proving \(1/n^2\) Tends to Zero
Let \(a_n=1/n^2\). We seek an index bound that makes \(|a_n-0|<\varepsilon\). Since \(n\) is positive,
Here \(\sqrt{\varepsilon}\) is positive, so taking the positive square root gives the last equivalent condition. For any \(\varepsilon>0\), choose
Then \(N>1/\sqrt{\varepsilon}\). If \(n\geq N\), it follows that \(n>1/\sqrt{\varepsilon}\). Both sides are positive, so squaring preserves the inequality and gives \(n^2>1/\varepsilon\). Taking reciprocals yields \(1/n^2<\varepsilon\).
This verifies the required estimate for every \(n\geq N\), not merely at \(N\). Since \(\varepsilon>0\) was arbitrary, the definition proves that \(1/n^2\) converges to \(0\). The chosen bound checks directly: \(n\geq N>1/\sqrt{\varepsilon}\) implies \(n^2>1/\varepsilon\), and therefore \(1/n^2<\varepsilon\).
Why the Method Works
The method succeeds because it reverses the direction of a proof without changing its logical requirement. The definition asks for an index after which the error is below \(\varepsilon\). Algebra tells us a condition on \(n\) that is sufficient for that error estimate. Choosing \(N\) to meet that condition ensures all larger indices meet it too, provided the expression and inequalities used give that implication.
Take an arbitrary \(\varepsilon>0\). The choice must be valid for every such tolerance.
Calculate \(|a_n-L|\), where \(L\) is the proposed limit.
Find a condition on \(n\) that guarantees \(|a_n-L|<\varepsilon\).
Pick \(N\in\mathbb{N}\) so that every \(n\geq N\) satisfies the condition.
Starting from \(n\geq N\), show explicitly that \(|a_n-L|<\varepsilon\), then apply the definition.
A choice of \(N\) is a sufficient bound, not necessarily an optimal one. If one choice makes the proof work, there is no need to minimize it. Sometimes an easier estimate gives a larger \(N\) than exact algebra would give. That is acceptable as long as \(N\) is a natural number and the estimate holds for every \(n\geq N\).
A Common Mistake: Rounding the Wrong Way
The condition \(n>1/\varepsilon\) calls for an integer strictly greater than the threshold. Simply taking the floor may fail. For instance, with \(\varepsilon=0.4\), the threshold is \(1/\varepsilon=2.5\). Taking \(N=\lfloor 2.5\rfloor=2\) does not work for \(a_n=1/n\): at \(n=2\),
The correction is to choose \(N=\lfloor 2.5\rfloor+1=3\). Then every \(n\geq3\) satisfies \(1/n\leq1/3<0.4\). More generally, \(\lfloor x\rfloor+1>x\), so it handles a strict threshold even when \(x\) is already an integer.
What to Check in Your Own Proof
A direct epsilon-N proof is complete only when its choices and quantifiers are explicit. Check that \(\varepsilon\) was arbitrary and positive, that \(N\) is a natural number, and that \(N\) depends only on \(\varepsilon\) and fixed quantities in the sequence—not on the later index \(n\). Then begin with an arbitrary \(n\geq N\) and derive the required strict error inequality.
It is also important to keep track of strict and non-strict inequalities. The definition requires \(|a_n-L|<\varepsilon\). An intermediate estimate such as \(|a_n-L|\leq\varepsilon\) does not by itself establish that requirement. A strict inequality in the index bound, or a stronger estimate, should deliver the strict conclusion.
These proofs make convergence concrete: for each requested accuracy, they give a point in the sequence after which every term meets that accuracy. Solving for the index is the practical bridge between an error formula and the formal definition.
Check Your Understanding
For each question, pay attention to the index choice and verify the final inequality from the condition \(n\geq N\).
- For \(a_n=2/n\), what condition on \(n\) ensures \(|a_n-0|<\varepsilon\), and what natural-number choice of \(N\) works?
- Why must the index \(N\) be chosen after \(\varepsilon\) is given?
- For \(a_n=6+1/(n+2)\), write the error from \(6\) and give a sufficient choice of \(N\).
- For \(a_n=1/n^2\), explain why \(n>1/\sqrt{\varepsilon}\) implies the required error estimate.
- With \(\varepsilon=0.2\), why is \(N=\lfloor1/\varepsilon\rfloor\) not a valid choice for the sequence \(1/n\)?