convergence of sequence of averages the other way around In a vector normed space, if $ \{x_n\} \longrightarrow x $ then  $ z_n = \dfrac{x_1 + \cdots+x_n}{n} \longrightarrow x $ 
Is it true the other way arround too? meaning:  if $ z_n = \dfrac{x_1 + \cdots+x_n}{n} \longrightarrow x $ then   $ \{x_n\} \longrightarrow x $?
My intuition says that it is true because $ z_n = \dfrac{x_1 + \cdots+x_{n-1}}{n} + \frac{x_n}{n}\longrightarrow x$ 
Thanks :)
 A: Your question is,

If the average of the first $n$ terms of a sequence tends to a limit, does the sequence itself tend to a limit?

The answer is no in general, as is discussed in the comments.  The simplest counterexamples are the sequences which oscillate between two different values $\alpha$ and $\beta$; we would expect that the average of the first $n$ terms of such a sequence will tend to the average of $\alpha$ and $\beta$.  As a concrete example let's define
$$
x_n = \frac{1+(-1)^n}{2},
$$
so that $x_n$ alternates between $0$ (when $n$ is odd) and $1$ (when $n$ is even).  We then have
$$
n \, z_n = x_1 + x_2 + \cdots + x_n = \begin{cases}
\frac{n}{2} & \text{if } n \text{ is even}, \\
\frac{n-1}{2} & \text{if } n \text{ is odd},
\end{cases}
$$
from which we can deduce that
$$
\frac{1}{2} - \frac{1}{2n} \leq z_n \leq \frac{1}{2}
$$
and thus
$$
\lim_{n \to \infty} z_n = \frac{1}{2}.
$$
There are, however, many cases where one can deduce the convergence of the original sequence from the convergence of this average.
For example, if $(x_n)$ is a positive, monotonic sequence, then the convergence of $(z_n)$ implies the convergence of $(x_n)$.
A slightly more difficult (and more useful) example is discussed in this thread.
Results of this general shape are called Tauberian theorems.  A nice reference is the book Divergent Series by G. H. Hardy.
