Two sequences which have an average tend to zero 
Let $(a_n)_{n\geq 1}, (b_n)_{n\geq 1}$ be two sequence of positive real numbers such that $$ \lim_{n\to +\infty}\frac{a_1+a_2+\cdots+a_n}{n}=\lim_{n\to +\infty}\frac{b_1+b_2+\cdots+b_n}{n}=0.$$


Conjecture. For all $\epsilon>0$, there are infinitely many values of indices $k$ such that $a_k<\epsilon$ and $b_k<\epsilon.$

I think that this is true but I can not prove it now.
In the special case where $a_n = b_n$, that is, there is only one sequence, then one can argue easily using a contradiction argument. In the general case, the hard part is to show that the same set of indices are shared by both sequences $(a_n)_{n\geq 1}, (b_n)_{n\geq 1}$.
 A: The sequence $(c_n)$ defined by $c_n = a_n + b_n$ satisfies
$$
\lim_{n\to \infty}\frac{c_1+c_2+\cdots+c_n}{n}=0
$$
as well. It follows that 
$$
 \liminf_{n \to \infty} c_n = 0
$$
and in particular, for every $\epsilon > 0$,
$$
 \max(a_k, b_k)  \le c_k < \epsilon
$$
for infinitely many $k$.
A: Suppose by contradiction that exists $\varepsilon>0$ and $N \in \mathbb{N}$ such that $c_n:=a_n+b_n\geq\varepsilon$, for $n>N$. Then
$$
\frac{c_N+\dotsc+c_n}{n}\geq\frac{(n-N+1)}{n}\varepsilon \to \varepsilon,
$$
which is a contradiction. Therefore there exists infinitely many $k$s such that $a_k,b_k<c_k<\varepsilon$.
A: Let:
$$
F(n) = {\sum_{i=1}^n{a_i}\over n}\\
G(n) = {\sum_{i=1}^n{b_i}\over n}
$$
The conditions above are that
$$
\lim_{n\to\infty}F(n) = \lim_{n\to\infty}G(n) = 0
$$
Let $p(n,\epsilon)$ be the fraction of $a_i$ that are $\geq \epsilon$, for $i$ up to $n$.  Likewise $q(n,\epsilon)$ be the fraction of $b_1 \geq \epsilon$ for $i \leq n$.
Derive $F'(n,\epsilon)$ as follows: round each $a_i$ down to zero if $a_i < \epsilon$ or to $\epsilon$ if $a_i \geq \epsilon$.  Clearly, since each term in this sum is less than or equal to the sum in $F(n)$, it must always be the case that 
$$\forall \epsilon > 0: F'(n,\epsilon) \leq F(n)$$
Also, it is straightforward to calculate:
$$
F'(n,\epsilon) = p(n,\epsilon) \times \epsilon
$$
Thus, we can conclude that since the limit of $F(n)$ is zero, so must also be the limit of $F'(n,\epsilon)$, and since $\epsilon$ does not vary with $n$:
$$
\forall \epsilon: \lim_{n\to\infty}p(n,\epsilon) = 0$$
By the definition of limit (and the fact that $1/2 > 0$), there must exist an M such that for all $n > M$, $p(n,\epsilon) < 1/2$.
Likewise for $q(n,\epsilon)$; there must be an $M'$ such that for all $n > M'$, $q(n,\epsilon) < 1/2$.
The minimum possible overlap of cases where $a_i < \epsilon$ and $b_i < \epsilon$ is:
$$r(n,\epsilon) = 1 - p(n,\epsilon) - q(n,\epsilon)$$
Let $T = \max(M,M')$. Clearly, for all $n > T$: $r(n,\epsilon) > 0$.  Thus, we have an infinite set of numbers of which a positive fraction must have $a_i < \epsilon$ and $b_i < \epsilon$.  Thus, there must be an infinite number of such cases.
A: Well, one way to think about it:
As $\lim \frac{\sum a_i}n \to 0$ and $\lim \frac{\sum b_i}n \to 0$ then $\lim (\frac{\sum a_i}n + \frac{\sum b_i}n) \to 0+ 0 = 0$.
And $(\frac{\sum a_i}n + (\frac{\sum a_i}n = \frac {\sum_{i=0}^n (a_i + b_i)}n$ so
$\lim_{n\to \infty}(\frac{\sum a_i}n) + \lim_{n\to \infty}(\frac{\sum a_i}n) = \lim_{n\to \infty}\frac {\sum_{i=0}^n (a_i + b_i)}n =0$.
(Note:  We CAN'T do this with sums that do not converge.)
Now for $c_i = a_i + b_i$ for any $\epsilon > 0$, I claim there are infinite many $c_i < \epsilon$.
Why?
Because if there are were only finitely many then there would be a last $c_N < \epsilon$ (or no $c_i < \epsilon$ at all) all for all $k > N$ then $c_k \ge \epsilon$. If $\sum_{i=0}^N c_i = K$ then $\sum_{i=0}^{n; n > N} c_i = K + \sum_{i=N+1}^n c_i \ge K + (n-N)\epsilon$.
So $\frac {\sum_{i=0}^n c_i}n \ge \frac Kn + \frac (n-N){n}\epsilon = \frac Kn -\frac Nn\epsilon + \frac nn\epsilon = \frac {K-N\epsilon}n + \epsilon$.
So $\lim_{n\to \infty} \frac {\sum_{i=0}^n c_i}n \ge \lim_{n\to \infty}\frac {K-N\epsilon}n + \epsilon = \epsilon > 0$.
That's a contradiction so there are infinitely many $c_n < \epsilon.
.....
And that mean there are infinitely many $a_k + b_k < \epsilon$ and as each $a_k, b_k$ is positive $a_k < a_k +b_k < \epsilon$ and $b_k < a_k + b_k < \epsilon$.
