A series $\sum a_k$ is a sequence in disguise: its partial sums $s_n = a_1 + \cdots + a_n$. You met partial sums in Calculus 2. Since we rarely know the sum of a series in advance, the right tool is the Cauchy condition, which talks only about the terms.
🔑 Key idea
$\sum a_k$ converges to $S$ when $s_n \to S$. Cauchy criterion: $\sum a_k$ converges if and only if for every $\varepsilon > 0$ there is an $N$ with $|a_{m+1} + a_{m+2} + \cdots + a_n| < \varepsilon$ whenever $n > m \ge N$. Taking $n = m + 1$ shows that a convergent series has $a_k \to 0$.
💡 Why it works
The block $a_{m+1} + \cdots + a_n$ is exactly $s_n - s_m$. So "far-out blocks are small" is the statement "the partial sums form a Cauchy sequence," which in $\mathbb{R}$ is the same as "the partial sums converge." The criterion is useful because blocks can often be estimated even when the sum is unknown, as in $\sum \frac{1}{k^{2}}$: comparing with the telescoping $\frac{1}{k - 1} - \frac{1}{k}$ gives $s_n - s_m < \frac{1}{m}$, which is small for large $m$. So $\sum \frac{1}{k^{2}}$ converges, and we never needed its sum ($\frac{\pi^{2}}{6}$).
The same criterion explains the classic divergence. For the harmonic series, the block $\frac{1}{n + 1} + \cdots + \frac{1}{2n}$ has $n$ terms, each at least $\frac{1}{2n}$, so it is at least $\frac{1}{2}$ no matter how far out it starts. The criterion fails for $\varepsilon = \frac{1}{2}$, so the series diverges even though its terms go to $0$.
Partial sums of $\sum \frac{1}{k^{2}}$ settle down; switch to the harmonic series to see sums that keep creeping up.
Worked examples
✏️ Worked example: A telescoping series
Find $s_n$ and the sum for $\sum_{k=1}^{\infty} \frac{1}{k(k + 1)}$, and the smallest $n$ with $S - s_n < 0.01$.
$\frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}$, so the sum telescopes: $s_n = 1 - \frac{1}{n + 1}$.
$s_n \to 1$, so $S = 1$ and $S - s_n = \frac{1}{n + 1}$.
$\frac{1}{n + 1} < 0.01$ means $n + 1 > 100$, so $n \ge 100$.
For $\sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{k} = 1$, how many terms make the error less than $0.001$?
The tail after $n$ terms is $\sum_{k > n} \left(\frac{1}{2}\right)^{k} = \left(\frac{1}{2}\right)^{n}$.
$\frac{1}{2^{n}} < 0.001$ means $2^{n} > 1000$, so $n \ge 10$ ($2^{10} = 1024$).
Answer: $10$ terms.
⚠️ Common mistake
"The terms go to $0$, so the series converges." That is the most common series error. $a_k \to 0$ is necessary (it is the Cauchy criterion with blocks of one term), but not sufficient: $\sum \frac{1}{k}$ has terms going to $0$ and diverges, because long blocks do not get small. Use $a_k \to 0$ only to prove divergence: if the terms do not go to $0$, the series diverges.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.