Textbook
A sequence $a_1,a_2,\ldots$ has limit $L$ if, for every $\varepsilon>0$, some index $N$ ensures $|a_n-L|<\varepsilon$ for all $n\ge N$. The notation is $a_n\to L$ as $n\to\infty$.
The index $N$ may depend on the accuracy $\varepsilon$. Beyond it, the entire tail must satisfy the bound, not just infinitely many selected terms. Any finite initial segment does not affect existence or value of the limit.
For $a_n=1/n$, the limit is 0. Given any $\varepsilon>0$, there exists a natural $N>1/\varepsilon$; then for $n\ge N$, $|a_n|=1/n\le1/N<\varepsilon$. No term equals zero, yet the limit exists.
The alternating sequence $(-1)^n$ has no limit: its terms keep returning to −1 and 1. One convergent subsequence is not enough. Unlike a finite list, an infinite tail cannot be checked by enumerating every term.
The definition fixes the order of requirements: first an arbitrary positive tolerance is chosen, then a suitable index. One index need not work for all tolerances simultaneously. Convergence also does not require monotonicity: terms may approach from both sides provided every sufficiently late term remains inside each requested corridor. A proof handles an arbitrary tolerance; checking several numerical tolerances illustrates that proof rather than replacing it.
Definitions and notation: Basic Analysis I.
Related concepts: Function limits: making “approaches” precise.
In plain language
A sequence limit resembles a corridor containing the entire queue of approximations from some point onward. Each narrower corridor may require a later starting point, but one well-behaved visitor promises nothing about the rest. The tail of $1/n$ approaches zero even though zero need not present a passport as a queue member. Motion may oscillate; the requirement concerns the whole tail, not improvement at every step or the existence of a final term.
Example
For the constructed sequence $1/n$, the required error is strictly below 0.1. The example checks the boundary and a few tail terms.
Limits of sequences
Beginning
The first term has error 1 relative to zero.
Step 1 of 4: Beginning
Read all steps
Beginning
The first term has error 1 relative to zero.
- n; value; tolerance
- 11excluded0.1
The boundary is insufficient
At n = 10 the error equals 0.1, but the requirement is strict.
- n; value; tolerance
- 100.1excluded0.1current
Choice of N
Term 11 is below 0.1; N = 11 is valid.
- n; value; tolerance
- 11current1/11included0.1
The entire tail
Later denominators are at least 11, so every later error remains below the tolerance.
- Indices
- 1120100
- Values
- 1/11included0.05included0.01included
For tolerance 0.1, index 11 works, but it is not the only choice: 12 or 100 also starts a valid tail. The definition does not require the smallest index. Reducing the tolerance to 0.01 defeats the earlier choice and requires an index greater than 100. In both cases, the same inequality $1/n\le1/N$ covers every later term at once. That bound turns the displayed observation into a checkable argument.
The crucial phrases are “for every tolerance” and “for all indices beyond N.”