In Calculus 2 you said $a_n \to L$ when the terms "get closer and closer" to $L$. That phrase is not precise enough to prove anything: $1 + \frac{1}{n}$ gets closer and closer to $0$ as well, without ever getting close. The fix is a definition with two quantifiers, the kind you read in Discrete Math, and it rests on the Archimedean property.
Key idea
$a_n \to L$ means: for every $\varepsilon > 0$ there is an $N$ such that $|a_n - L| < \varepsilon$ for every $n \ge N$. The $\varepsilon$ is a challenge (how close?), and $N$ is the answer (from where on?). If one $N$ works, every larger $N$ works too.
Why it works
Read the definition as a game. A skeptic names a tolerance $\varepsilon$, as tiny as they like. You must answer with a point $N$ in the sequence beyond which every term stays within $\varepsilon$ of $L$, not just one term, and not just most of them. A sequence converges to $L$ when you can always win, whatever $\varepsilon$ the skeptic picks. For $1 + \frac{1}{n}$ and $L = 0$ the skeptic wins with $\varepsilon = \frac{1}{2}$: every term is more than $1$ away from $0$.
Worked examples
Worked example: Finding $N$ for one $\varepsilon$
Let $a_n = \frac{2n + 1}{n + 3}$, so $L = 2$. Find an $N$ that works for $\varepsilon = 0.01$.
- $a_n - 2 = \frac{2n + 1 - 2(n + 3)}{n + 3} = \frac{-5}{n + 3}$, so $|a_n - 2| = \frac{5}{n + 3}$.
- $\frac{5}{n + 3} < 0.01$ means $n + 3 > 500$, that is, $n > 497$.
- Since $\frac{5}{n + 3}$ decreases, every $n \ge 498$ works, and $n = 497$ gives exactly $0.01$, which is not less than $0.01$.
Answer: $N = 498$ (or any larger $N$).
Worked example: Checking a proposed $N$
For $a_n = \frac{1}{n^{2}}$ ($L = 0$) and $\varepsilon = 0.001$, does $N = 30$ work?
- $|a_n - 0| = \frac{1}{n^{2}} < 0.001$ means $n^{2} > 1000$, that is, $n \ge 32$.
- At $n = 30$ the distance is $\frac{1}{900} \approx 0.0011$, which is not below $0.001$.
Answer: No. The smallest $N$ that works is $32$.
Worked example: A sequence that zigzags
Let $a_n = \frac{3 + 2(-1)^{n}}{n}$, with limit $0$. Find the smallest $N$ that works for $\varepsilon = 0.1$.
- For even $n$, $|a_n| = \frac{5}{n}$; for odd $n$, $|a_n| = \frac{1}{n}$. The distances are not monotone.
- Even $n$: need $\frac{5}{n} < 0.1$, so $n > 50$; the last even failure is $n = 50$. Odd $n$: $\frac{1}{n} < 0.1$ once $n > 10$; the last odd failure is $n = 9$.
- The last failing index is $50$, so every $n \ge 51$ works.
Answer: $N = 51$.
Common mistake
Checking one term instead of all the later ones. In the zigzag example, $|a_{11}| = \frac{1}{11} < 0.1$, but $N = 11$ fails because $a_{12}, a_{14}, \ldots$ are still too far. Always ask: does the inequality hold for every $n \ge N$? The opposite slip is thinking there is one correct $N$. If $N = 498$ works, so do $499$ and $10^{6}$; a definition that says "there is an $N$" never needs the smallest one.