What is the difference between Cauchy and convergent sequence? I am really confused. I will appreciate if somebody can help me to define the difference between Cauchy and convergent sequence. 
Many thanks.
 A: A convergent sequence is also a Cauchy sequence.
A Cauchy sequence is not necessarily a convergent sequence. For example if our space is $X=\mathbb Q$, then 
$$
x_n=\frac{\lfloor n\sqrt{2}\rfloor}{n},
$$
is a Cauchy sequence which DOES NOT converge is $\mathbb Q$. It DOES converge is $\mathbb R$ but not in $\mathbb Q$. 
A metric space where every Cauchy sequence converges is called complete. The space $\mathbb Q$ with metric the distance $|a-b|$ is not complete, while $\mathbb R$ with the same metric is complete. 
A: Every convergent sequence is a cauchy sequence. The converse may however not hold. For sequences in $\mathbb{R}^k$ the two notions are equal. More generally we call an abstract metric space $X$ such that every cauchy sequence in $X$ converges to a point in $X$ a complete metric space. One can for instance show that the space of continuous functions on a compact set with the uniform metric is complete. 
As an example of a space that is not complete, consider the interval $(0,1)$ and the sequence $1/n$, clearly $1/n$ is cauchy, but $1/n$ is not convergent in $(0,1)$ as $0$ is not in $(0,1)$.
In contrast $[0,1]$ is complete, since it is a closed subset of the complete space $\mathbb{R}$.
A: Let's start in metric space $\mathbb{R}$ equipped with its normal
metric. 
A sequence $\left(x_{n}\right)$ in $\mathbb{R}$ is convergent
if some $x\in\mathbb{R}$ exists such that for every $\varepsilon>0$
some $n_{0}\in\mathbb{N}$ can be found such that $\left|x-x_{n}\right|<\varepsilon$
for each $n\geq n_{0}$. If that is the case then it can be shown
that this $x$ is unique and the sequence is said to converge to $x$.
A sequence $\left(x_{n}\right)$ in $\mathbb{R}$ is a Cauchy-sequence
if for every $\varepsilon>0$ some $n_{0}\in\mathbb{N}$ can be found
such that $\left|x_{n}-x_{m}\right|<\varepsilon$ for each pair $n,m\in\mathbb{N}$
with $n,m\geq n_{0}$. 
It can be shown that a convergent sequence is a Cauchy-sequence. This
in any metric space.
In the case we are dealing with (metric space $\mathbb{R}$) the opposite
is also true: every Cauchy-sequence in $\mathbb{R}$ is a convergent
sequence.
However, there are metric spaces in wich the opposite is not true.
An example is $\mathbb{Q}$. Start with a sequence $\left(x_{n}\right)$
in $\mathbb{R}$ that converges to some $x\in\mathbb{R}-\mathbb{Q}$
and let it be that $x_{n}\in\mathbb{Q}$ for each $n$. It is a Cauchy-sequence
in $\mathbb{R}$ and is convergent in $\mathbb{R}$. However, if we look at
it as a sequence in $\mathbb{Q}$ then it is still a Cauchy-sequence, but it is not convergent, simply
because $x\notin\mathbb{Q}$.
Spaces in which the opposite is true are so-called 'complete spaces'.
A: Let $(X,d)$ be a metric space.

Definition. A sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in X$ for all $n\in\mathbb N$ is a Cauchy sequence in $X$ if and only if for every $\varepsilon > 0$ there exists $N\in\mathbb N$ such that $d(x_n,x_m)<\varepsilon$ for all $n,m>N$.

Informally speaking, a Cauchy sequence is a sequence where the terms of the sequence are getting closer and closer to each other.

Definition. A sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in X$ for all $n\in\mathbb N$ is convergent if and only if there exists a point $x\in X$ such that for every $\varepsilon > 0$ there exists $N\in\mathbb N$ such that $d(x_n,x)<\varepsilon$. In this situation we say $x$ is a limit of the sequence $(x_n)_{n\in\mathbb N}$, or $(x_n)_{n\in\mathbb N}$ converges to $x$.

Informally speaking, a sequence is convergent if the terms of the sequence are getting closer and closer to some point $x\in X$.
A metric space $X$ is said to be complete if every Cauchy sequence is convergent. This is the case for the spaces $\mathbb R^n$, which is the reason why you might not see the difference of the concepts at first glance.
Let's take a look at a familiar metric space which is not complete, so we have Cauchy sequence that are not convergent. Let $X=\mathbb Q$ be the set of rational numbers with the usual metric given by $d(x,y)=\left|x-y\right|$. Consider the sequence defined by
$$
x_1 = 1,\quad
x_{n+1} = 1+\frac{1}{1+x_n}.
$$
We can show that this sequence is a Cauchy sequence in $\mathbb Q$, but there is no $x\in \mathbb Q$ that is a limit of the sequence.
Why is that? Well, consider the same sequence as a sequence in the complete metric space $X=\mathbb R$, we can now show that $(x_n)_{n\in\mathbb N}$ converges to $\sqrt 2\in\mathbb R$, but $\sqrt 2$ is not in $\mathbb Q$. So while the sequence converges in $\mathbb R$, it doesn't converge in $\mathbb Q$. Nevertheless it is a Cauchy sequence in both spaces.
A: When you have a convergent sequence: $x_n\to\xi$, then for large $n$ all points $x_n$ are near $\xi$, and by the triangle inequality they are then near each other as well – in formal language: The $x_n$ automatically form a Cauchy sequence; no big deal.
But:
In some metric spaces $X$, among them ${\mathbb R}$, ${\mathbb R}^n$, and ${\mathbb C}$, the converse also holds: A Cauchy sequence is automatically convergent. (Such spaces are called complete.) This property of spaces is all-important for the following reason: In order to test a sequence $(x_n)_{\geq1}$ for Cauchy-ness you only have to look at the given $x_n$ themselves, and you don't have to know the limit beforehand. When the sequence passes this "finitary" test you know for sure that it has a limit $\xi\in X$. 
In all, the so-called Cauchy criterion (for real sequences, say) is not a little proposition about convergence, but a deep theorem about the fine structure of ${\mathbb R}$.
