Definition of convergence of a series My text book gives the following definition of convergence of a series:

A sequence of real numbers is said to be convergent if it has a finite real number $L$ as its limit. We then say that the sequence $x$ converges to $L$.

But I find it to be confusing, as from the above definition of convergence, $1/n$ turns out to be a Cauchy sequence.
But wikipedia seems to be providing a different definition of convergence Definition of Convergent Series
PS: I am just a newbie in real analysis, so I do request you to be a little more elaborative.
Edit: Oh, that question was stupid enough. I really did mix up the two terms "sequence" and "series". Thanks for pointing this out.

There are two different kinds of objects that you are studying:  sequences, and series.
A sequence $(a_n)$ of real numbers converges if there is some finite real number $L$ such that $\lim_{n\to\infty} a_n = L$.  What this means is that we can make the difference between $a_n$ and $L$ as small as we like by choosing a number $n$ that is large enough.  More formally,

A sequence $(a_n)$ converges to $L$ if for any $\varepsilon > 0$ there exists some $N$ so large that $n \ge N$ implies that $|a_n - L| < \varepsilon$.

A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $S_N$ converges, where
$$ S_N := \sum_{n=1}^{N} a_n. $$
That is, in order to discuss the convergence of a series, we first turn the series into a sequence, then seek to understand the properties of that sequence.  Thus a series is said to converge to a limit $S$ if the sequence $(S_N)$ (as defined above) converges to $S$ as a sequence.  In notation, we might write
$$ \sum_{n=1}^{\infty} a_n = S \iff \lim_{N\to\infty} S_n = \lim_{N\to\infty} \left( \sum_{n=1}^{N} a_n \right) = S.$$
The classic example (cited by other responses to your question) is the harmonic series,
$$ \sum_{n=1}^{\infty} \frac{1}{n}. $$
The individual terms $\frac{1}{n} \to 0$ as $n\to \infty$, but the series does not converge.  To see this, write
$$ \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \dotsb.$$
Notice that $\frac{1}{3} > \frac{1}{4}$, thus
$$ \frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. $$
Similarly, each term in the next group is bigger than $\frac{1}{8}$, and so
$$ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}. $$
By a similar process, we can continue to lump terms together to add up to $\frac{1}{2}$, which gives
$$ \sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \dotsb \ge \sum_{k=1}^{\infty} \frac{1}{2} = \infty, $$
where the inequality can be justified via the Squeeze Theorem.
