# 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 :)

• If $x_n=(-1)^n$ then $x_n$ does not converge yet $z_n \to 1/2$. The topic is methods of summation, I think "Cesaro sum"... – coffeemath Mar 5 '13 at 17:10
• @coffeemath I think you mean $x_n = (1 + (-1)^n)/2$. – Antonio Vargas Mar 5 '13 at 17:50
• @Antonio Yes, I have the wrong limit. If $x_n=(-1)^n$ as I have it, then $z_n \to 0$. And with your adjustment $x_n=(1+(-1)^n)/2$ the limit is $z_n \to 1/2$. Any limit like this shows the "other way around" idea doesn't work. – coffeemath Mar 5 '13 at 18:02

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.