Two facts about convergence come straight from the definition: a convergent sequence cannot wander off to infinity, and a sequence can fail to converge in a precise way that you can exhibit with a single $\varepsilon$.
Key idea
Convergent sequences are bounded: if $a_n \to L$, some $M$ has $|a_n| \le M$ for every $n$. Not converging to $L$ means: there is an $\varepsilon > 0$ such that for every $N$ some $n \ge N$ has $|a_n - L| \ge \varepsilon$.
Why it works
Bounded. Use the definition with $\varepsilon = 1$: there is an $N$ with $|a_n - L| < 1$ for $n \ge N$, so those terms have $|a_n| < |L| + 1$ (triangle inequality). Only finitely many terms come before $N$. So $M = \max\{|a_1|, \ldots, |a_{N-1}|, |L| + 1\}$ bounds every term. $\blacksquare$ Negation. "For every $\varepsilon$ there is an $N$ such that for all $n \ge N$, $|a_n - L| < \varepsilon$" becomes, one quantifier at a time: there is an $\varepsilon$ such that for every $N$ there is an $n \ge N$ with $|a_n - L| \ge \varepsilon$.
Worked examples
Worked example: The bound from the proof
For $a_n = \frac{5n + 9}{n + 1}$, which tends to $5$, carry out the proof's recipe: find $N$ for $\varepsilon = 1$ and then $M$.
- $|a_n - 5| = \frac{4}{n + 1} < 1$ once $n + 1 > 4$, so the smallest $N$ is $4$ ($|a_3 - 5| = 1$ exactly, which is not less than $1$).
- Earlier terms: $a_1 = 7$, $a_2 = \frac{19}{3}$, $a_3 = 6$. And $|L| + 1 = 6$.
- $M = \max\{7, \frac{19}{3}, 6, 6\} = 7$.
Answer: $N = 4$, $M = 7$.
Worked example: Proving divergence with one $\varepsilon$
Show that $a_n = (-1)^{n}$ does not converge to $L = \frac{1}{2}$, and give the largest $\varepsilon$ that witnesses it.
- The terms are $-1$ and $1$, each infinitely often. Their distances from $\frac{1}{2}$ are $\frac{3}{2}$ and $\frac{1}{2}$.
- Beyond any $N$ there is an odd $n$, where $|a_n - \frac{1}{2}| = \frac{3}{2}$.
- So every $\varepsilon$ with $0 < \varepsilon \le \frac{3}{2}$ witnesses the failure; $\frac{3}{2}$ is the largest.
Answer: $\varepsilon = \frac{3}{2}$ (or anything smaller and positive). Since $L$ was arbitrary in this style of argument, $(-1)^{n}$ diverges.
The contrapositive of "convergent implies bounded" is a quick test for divergence: an unbounded sequence diverges. So $\sqrt{n}$, $\ln n$ and $(-2)^{n}$ all diverge, with no $\varepsilon$ needed. A bounded sequence needs a closer look.
Common mistake
Two traps. First, "bounded implies convergent" is false: $(-1)^{n}$ is bounded and diverges. Only the converse is true. Second, negating the definition as "for every $\varepsilon$, no $N$ works" says far too much: it would mean the sequence is never within any distance of $L$. The negation needs just one bad $\varepsilon$, and then for that $\varepsilon$ every $N$ fails somewhere later.