Divergence Test
Summary
If the series $\sum\limits_{n = 1}^{\infty} a_{n}$ converges, then the sequence $\{a_{n}\}$ converges to $0$. $$ \sum\limits_{n = 1}^{\infty} a_{n} \text{ is convergent } \implies \lim\limits_{n \to \infty} a_{n} = 0 $$
Proof
Let the sum of the series be $\sum\limits_{n = 1}^{\infty} a_{n} = s$. That is, for the partial sum $s_{n}$, it is $\lim\limits_{n \to \infty} s_{n} = s$. Then, since $a_{n} = s_{n} - s_{n-1}$,
$$ \lim\limits_{n \to \infty} a_{n} = \lim\limits_{n \to \infty} (s_{n} - s_{n-1}) = \lim\limits_{n \to \infty} s_{n} - \lim\limits_{n \to \infty} s_{n-1} = s - s = 0 $$
■
Explanation
The converse is not true. In other words, if the sequence $\{a_{n}\}$ converges to $0$, it does not mean that the series $\sum\limits_{n = 1}^{\infty} a_{n}$ converges. A well-known example is the harmonic series. The harmonic sequence $\left\{ \dfrac{1}{n} \right\}$ converges to $0$, but the harmonic series does not converge.
$$ \lim\limits_{n \to \infty} \dfrac{1}{n} = 0 \quad \text{ but } \quad \sum\limits_{n = 1}^{\infty} \dfrac{1}{n} = \infty $$
The contrapositive is the divergence test.
Divergence Test
If the sequence $\{a_{n}\}$ does not converge to $0$, then the series $\sum\limits_{n = 1}^{\infty} a_{n}$ diverges.