To use the definition of a limit you must already know the limit $L$. Cauchy's idea removes $L$: a sequence should converge when its terms get close to each other. In $\mathbb{R}$ that idea is exactly right, thanks to Bolzano–Weierstrass, and it is how we will handle series, whose sums are usually unknown.
Key idea
$(a_n)$ is Cauchy when for every $\varepsilon > 0$ there is an $N$ with $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$. Theorem. A sequence of real numbers converges if and only if it is Cauchy.
Why it works
Convergent implies Cauchy: if $a_n \to L$, choose $N$ with $|a_n - L| < \frac{\varepsilon}{2}$ for $n \ge N$; then $|a_m - a_n| \le |a_m - L| + |L - a_n| < \varepsilon$. Cauchy implies convergent: a Cauchy sequence is bounded (use $\varepsilon = 1$ as for convergent sequences). By Bolzano–Weierstrass some subsequence $a_{n_k} \to L$. Given $\varepsilon > 0$, take $N$ from the Cauchy condition with $\frac{\varepsilon}{2}$, and some $n_k \ge N$ with $|a_{n_k} - L| < \frac{\varepsilon}{2}$. For $n \ge N$: $|a_n - L| \le |a_n - a_{n_k}| + |a_{n_k} - L| < \varepsilon$. $\blacksquare$ In $\mathbb{Q}$ the second half fails: the decimal truncations $1, 1.4, 1.41, 1.414, \ldots$ of $\sqrt{2}$ are Cauchy, with no rational limit.
Worked examples
Worked example: Finding $N$ for a Cauchy sequence
For $a_n = \frac{3}{n}$ and $\varepsilon = 0.1$, find the smallest $N$ with $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$.
- For $m, n \ge N$: $|a_m - a_n| = 3\left|\frac{1}{m} - \frac{1}{n}\right| < \frac{3}{N}$, and taking $m = N$ and $n$ huge comes as close to $\frac{3}{N}$ as you like.
- So every pair is within $0.1$ exactly when $\frac{3}{N} \le 0.1$, that is, $N \ge 30$.
Answer: $N = 30$ (or any larger $N$).
Worked example: Neighbors close, but not Cauchy
Is $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$ Cauchy?
- $H_{2n} - H_n = \frac{1}{n + 1} + \cdots + \frac{1}{2n}$ has $n$ terms, each at least $\frac{1}{2n}$, so it is at least $\frac{1}{2}$.
- For $\varepsilon = \frac{1}{2}$, no $N$ works (take $m = 2n$ with $n \ge N$).
Answer: No, so $H_n$ diverges, even though $H_{n+1} - H_n = \frac{1}{n + 1} \to 0$.
For a series $\sum a_k$, the difference of two partial sums is a block $a_{m+1} + \cdots + a_n$. So "the partial sums are Cauchy" says "far-out blocks are small," with no need to know the sum. That is the Cauchy criterion of the next unit.
Common mistake
Checking only consecutive terms. "$|a_{n+1} - a_n| \to 0$, so the sequence is Cauchy" is false: the harmonic sums above and $\sqrt{n}$ are counterexamples. The Cauchy condition is about every pair $m, n \ge N$, however far apart. Check with pairs like $(n, 2n)$ or $(n, 4n)$.