Prove that $\lim_{n\rightarrow \infty} (x_1+x_2+...+x_n)/n= \infty$ I need help with this exercise:
Consider the sequence $\{x_n\}$ that verifies $$\lim_{n\rightarrow \infty} x_n = +\infty$$
Now prove that $$\lim_{n\rightarrow\infty} \frac{x_1+x_2+...+x_n}n= +\infty$$
Also I need to find an example that the converse does not apply.
Thank you for all your answers.
 A: Let $M>0$. Since $\lim_{n\to\infty}x_n=\infty$, there exists $N>0$ such that
$$ x_n\ge 4M \text{ for }n> N. $$
For fixed $N$, since 
$$ \lim_{n\to\infty}\frac{x_1+x_2+\cdots+ x_{N}}{n}=0 $$
there is $N_1>N$ such that
$$ \frac{x_1+x_2+\cdots x_{N}}{n}+M>0. $$
So, for $n>\max\{2N,N_1\}$, one has
\begin{eqnarray} 
&&\frac{x_1+x_2+\cdots+ x_n}{n}\\
&=&\frac{x_1+x_2+\cdots+ x_{N}}{n}+\frac{x_{N+1}+x_{N+2}+\cdots x_{n}}{n} \\
&\ge&\frac{x_1+x_2+\cdots+ x_{N}}{n}+\frac{n-N}{n}\cdot4M\\
&\ge&\frac{x_1+x_2+\cdots+ x_{N}}{n}+\frac12\cdot4M\\
&\ge&M.
\end{eqnarray}
Thus
$$ \lim_{n\to\infty}\frac{x_1+x_2+\cdots+ x_n}{n}=\infty. $$
A: By definition, for any $N>0$ we need to show that there is $M$ such that for $n>M$ you have $\mu_n>N$. WLOG assume all the $x_i>0$ as a finite set of negatives won't affect the limit at all. Let $M',N'$ be the constants we have for $\{x_n\}$ (i.e. for all $n>M'$ we have $x_n>N'$) then so long as $n>M'$ we have
$$\mu_n={x_1+\ldots+x_n\over n}\ge {(n-M')N'\over n}$$
If we just go to $n>2M'$, then we have
$$\mu_n>{1\over 2}N'$$
which is what we needed to show.
The converse is clearly false, let $\{x_n\}$ be such that $x_{2n}=n$ and $x_{2n+1}=0$. Then $\mu_{2n}={n(n+1)\over 4n}={n+1\over 4}$ but clearly the limit of the $x_i$ does not exist.
