Theorem:
If the series $\sum_{k=1}^{\infty}a_k$ converges, then
$\displaystyle\lim_{n\to\infty}a_n=0$
Proof:
From the convergence of the sum, we know that the sequence of partial sums $\{S_n\}$ converges. Say $\displaystyle\lim_{n\to\infty}S_n=S$. From $a_n=S_n-S_{n-1}$, we will get: $\displaystyle\lim_{n\to\infty}a_n=\displaystyle\lim_{n\to\infty}(S_n-S_{n-1})=S-S=0$ Q.E.D
Now, is the opposite direction -- If $\displaystyle\lim_{n\to\infty}a_n=0$, then $\sum_{k=1}^{\infty}a_k$ converges, will work too? The answer is no!
Take the harmonic series, $\sum_{n=1}^{\infty}\frac{1}{n}$, as you can see $\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$, but is the $\sum_{n=1}^{\infty}\frac{1}{n}$ converges? No! And why not? Here:
First, Cauchy criteria:
$\sum_{k=1}^{\infty}a_k$ converges iff
$\forall\epsilon>0, \exists n, n>N(\epsilon), \forall p\in\mathbb{N}$ $|a_{n+1}+a_{n+2}+...+a_{n+p}|=|\sum_{k=n+1}^{n+p}a_k|<\epsilon$
Applying above on $\sum_{k=1}^{\infty}\frac{1}{k}$:
Take $p=n$,
$S_{2n}-S_n=\sum_{k=n+1}^{2n}\frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}>n \frac{1}{2n}=\frac{1}{2}$
Hence, $S_{2n}-S_n$ can't be equal less than given $\epsilon$. And therefor the series is diverges.