when a sequence is not cauchy does it mean that it is divergent? when a sequence is not Cauchy does it mean that it is divergent?
i have an example of a problem that shows that a sequence is not cauchy .i was wondering if i can usse it the same one to say that it means the sequence diverges too 
 A: Yes. It's divergent. We say that $\lim_{n \to \infty} a_{n} = \alpha$ if for every $\epsilon > 0$ we can pick $N \in \mathbb{N}$ such that if $n > N$, then $|a_{n} - \alpha| < \epsilon$, and it's Cauchy if for each such $\epsilon$ exists $M$ such that if $m, n > N$, then $|a_{m} - a_{n}| < \epsilon$. It's enough to show a convergent sequence is Cauchy. Pick $N'$ such that if $n > N$, then $|a_{n} - \alpha| < \frac{\epsilon}{2}$. Then
\begin{align*}
|a_{m} - a_{n}| & = |(a_{m} - \alpha) - (a_{n} - \alpha)| \\
& \leq |a_{m} - \alpha| + |a_{n} - \alpha| & ( \textrm{triangle inequality})\\
& < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\
& = \epsilon
\end{align*}
Thus a convergent sequence is Cauchy, and so a non-Cauchy sequence is not convergent.
A: Every convergent sequence is a Cauchy sequence, and in complete metric spaces, every Cauchy sequence converges.  The real line $\mathbb R$ and the complex plane $\mathbb C$ are complete metric spaces.
Here's a proof of the first assertion (which is the one you seem to need).  Suppose $a_n\to a$ as $n\to\infty$.  Then for every $\varepsilon>0$ there exists a natural number $N$ such that for every $n\ge N$ we have $|a_n-a|<\varepsilon/2$.  Consequently for every $n,m\ge N$ we have
$$
|a_n-a_m|=|(a_n-a)-(a_m-a)| \le |a_n-a|+|a_m-a| \le \frac \varepsilon 2+\frac \varepsilon 2 = \varepsilon.
$$
