Whats the difference between a series and sequence? I was looking at a question earlier that involved sequences and found out that the sequence converged to 0 but the series diverged to infinity. How is that possible? for example the sequence was $a_n$ $=$ $\frac{1}{x_n}$ 
The question was here: Convergence of $x_n=f\left(\frac{1}{n}\right), n\geq 1$ 

A series is in some sense a type of sequence. However, it is a sequence of partial sums.
For example, if we take some sequence $\{a_n\}_{n\geq 1}$, then we can in turn retrieve a series from this sequence by considering the following partial sums:
$S_N=a_1+...+a_n=\sum_{n=1}^{N}a_n$
Then if we consider the following sequence: $\{S_N\}_{n \geq 1}$ We have actually defined a series!
$S_1=a_1$
$S_2=a_1+a_2$
and so on.
Thus, the difference is the following:
Consider the sequence $\{a_n\}_{n \geq 1}$ defined by $a_n=\frac{1}{n}$.
It is intuitively clear (and if not, use the Archimedean property) that $\lim_{n \to \infty} \frac{1}{n}=0$.
Now, we can consider the sequence of partial sums:
Let $\{S_n\}_{n \geq 1}$ be defined by $S_n=1+\frac{1}{2}+...+\frac{1}{n}$
Then $$\lim_{n \to \infty} S_n=\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k}=\sum_{k=1}^{\infty}\frac{1}{n}$$
Which is called the harmonic series, and it diverges.
We can show this by either the integral test, or just note:
$\begin{align} &1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...\\
>&1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...\\
=&1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...
\end{align}$
Which very clearly diverges.
As an aside, it should be very clear that $\{S_n\}$ will not converge at all if $\{a_n\}$ does not, but the converse is not true.
If you want an example where they both converge, just take $\{a_n\}$ to be defined by $a_n=\frac{1}{2^n}$. Then you can consider the sequence of partial sums for $a_n$ (and hence, define a series.)
Then $\{a_n\} \to 0$ as $n \to \infty$. Yet we also have that 
$$\lim_{n \to \infty} S_n=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{1}{2^k}=\sum_{k=1}^{\infty}\frac{1}{2^k}=\frac{1/2}{1-1/2}=1$$
To see a derivation of the penultimate equality, you can look further into what are called geometric series
