Arithmetic mean sequence If the ssequence of $a_n$ goes to infinity as $n$ goes to infinity, then does $$\lim_{n\to\infty} \frac{a_1 + \dots + a_n}{n} = \infty?$$
I know this sequence converges to a finite value if the sequence $a_n$ converges to a finite value, but I don't know if that helps. I've tried using the definition a sequence converging to infinity, I've also tried using the convergence of $\frac{1}{a_n}$ to $0$ to show $\frac{n}{a_1 + \dots + a_n}$ converges to $0$, but no luck. Do I utilize the arithmetic mean inequality? Any hints are more than welcome (only hints please!).
 A: Indeed,
THEOREM If the limit of $a_n$ goes to infinity as $n$ goes to infinity, then $$\lim_{n\to\infty} \frac{a_1 + \dots + a_n}{n} = \infty$$
PROOF Let $\ C>0.\ $ There exists natural $\ N_C\ $ such that
$\ a_k>2\cdot C\ $ for every $\ k>N_C.\ $ Let
$$ B_C\ :=\ \sum_{n=1}^{N_C}\ a_n $$
and let natural $\ m_C\ $ satisfy
$$ m_C\ >\ N_C-\frac {B_C}C $$
so that
$$ B_C+2\cdot C\cdot m_C\ >\ (N_C+m_C)\cdot C $$
Now, let $\ n>N_C+m_C.\ $ Then
$$ \frac{a_1 + \dots + a_n}n\,\ >
    \,\ \frac{B_C\ +\ 2\cdot C\cdot m_C\ +\ \sum_{k=N_C+m_C+1}^n a_k}n $$
$$ >\,\ \frac{(N_C+m_C)\cdot C\,\ +\,\ (n-(N_C+m_C))\cdot2\cdot C}n
\,\ >\,\ C $$
Since $\ C>0\ $ is arbitrary, the theorem holds.   Great!
A: If the limit goes to infinity, then for any constant $C$. All terms bigger than some constant $N_C$ are bigger than $C$. Taking the arithmetic mean of the first $10N_C$ terms give you an arithmetic mean as large as $.9C$. Since I can make $C$ as large as I like, I can make $.9C$ as large as I like, so the arithmetic mean must go to infinity.
A: This might be a bit overkill but I think it is worth noting it:
Your question can be dealt with using the general form of the Cesaro-Stolz theorem:
$$+\infty=\lim_{n\to\infty}a_n=\liminf_{n\to \infty}\frac{a_n}{1} \leq \liminf_{n\to \infty}\frac{a_1 + \cdots + a_n}{n}$$
So, it follows immediately that $\boxed{\frac{a_1 + \cdots + a_n}{n}\stackrel{n \to \infty}{\longrightarrow}+\infty}$.
A: Let $S(n)=\frac {1}{n}(a_1+...a_n).$ Given $R>0,$ we want some $n_1$ such that $n\ge n_1\implies S(n)>R.$
First, take some $n_0\in \Bbb N$ such that $n> n_0\implies a_n>2R.$ Now if $n>n_0$ then $$S(n)=\frac {1}{n}(a_1+..+a_{n_0})+ \frac {1}{n}(a_{n_0+1}+...+a_n)>\frac {1}{n}(a_1+...+a_{n_0})+\frac {(n-n_0)}{n}\cdot 2R=T(n).$$ With $n_0$ fixed, we have $\lim_{n\to \infty}\frac {1}{n}(a_1+...+a_{n_0})=0$ and $\lim_{n\to \infty}\frac {(n-n_0)}{n}\cdot 2R=\lim_{n\to \infty}(1-\frac {n_0}{n})\cdot 2R=2R.$ Therefore $\lim_{n\to \infty}T(n)=2R.$
So there exists $n_1>n_0$ such that $$n\ge n_1\implies |T(n)-2R|<R/2\implies R<3R/2<T(n)<S(n)\implies R<S(n).$$
A: Intuitively, as $a_n$ goes to infinity, $a_n>2M$ finishes to hold for any $M$, how ever large. This means that the average $\overline{a_n}>M$ finishes to hold as well, because it is larger than the average of a finite sum and as many $2M$ terms as you want.
