Show that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\geq 0$？ Suppose $a_n$ is a sequence such that $a_n \downarrow 0$, 
 $\epsilon_n=-1$ or $1$ for all $n$, the series $\sum a_n$ diverges but the series $\sum_{n=1}^\infty \epsilon_n a_n$ converges. Is it true that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\geq 0$?

Question history: Originally, this question is about the assertion $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}=0$, which has been proved wrong by some wonderful answers. 

Note $a_n$ is non-increasing to zero here. Some people forget this assumption.  
 A: The statement is not true. Using Michael's construction, but simplified gives:
$$\epsilon_n = \{1, -1, -1, 1, 1, 1,1, \dots\}$$
$$\epsilon_n = (-1)^{\lfloor \log_2 n\rfloor}$$
After $1$, $7$, $31$, $\dots$, $2^{2k+1} - 1$ numbers we hit a maximum. This maximum is exactly:
$$\sum_{i=0}^{2k} (-2)^i = \frac{1}{3}(2^{2k+1} + 1)$$
So at this maximum our $\sum e_n/n = \frac{1}{3}\dfrac{2^{2k+1} + 1}{2^{2k+1} - 1}$. So for large $n$ our $\limsup \sum e_n/n = \frac{1}{3}$.

$$a_n = \{\frac1{1\cdot 2^0}, \frac1{2\cdot 2^1}, \frac1{2\cdot 2^1}, \frac1{3\cdot 2^2}, \frac1{3\cdot 2^2}, \frac1{3\cdot 2^2}, \frac1{3\cdot 2^2}, \frac1{4\cdot 2^3}, \dots\}$$
$$a_n = \{\frac11, \frac1{4}, \frac14, \frac1{12}, \frac1{12}, \frac1{12}, \frac1{12}, \frac1{32}, \dots\}$$
$$a_n = \frac{1}{(\lfloor\log_2 n\rfloor + 1)2^{\lfloor\log_2 n\rfloor}}$$
$a_n$ is the harmonic series, with the $k$th element repeated $2^{k-1}$ times, divided by $2^{k-1}$. $\sum a_n$ diverges because the harmonic series does. Then, $\epsilon_n a_n$ is the alternating harmonic series, with the $k$th element repeated $2^{k-1}$ times, divided by $2^{k-1}$. The sum of this series converges just like the alternating harmonic series does:
$$\sum_{n=1}^\infty \epsilon_na_n = \sum_{k=0}^\infty 2^{k}\cdot (-1)^{k} \cdot \frac{1}{(k + 1)2^{k}} = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \ln(2) \tag*{$\blacksquare$}$$
A: With these hypotheses it is true that $\limsup (\epsilon_1 + \ldots + \epsilon_n)/n \geqslant 0.$
Let $S_n = \epsilon_1 + \ldots + \epsilon_n$.  Summing by parts , we have
$$\sum_{j=1}^n \epsilon_ja_j = S_n a_n + \sum_{j=1}^{n-1} S_j(a_j - a_{j+1}).$$
If $\limsup S_n/n  =-r < 0$, then there exists $N$ such that $S_n/n < -r/2$ for all $n > N$ and
$$\tag{*}\sum_{j=1}^n \epsilon_ja_j < -\frac{r}{2}na_n + \sum_{j=1}^{N} S_j(a_j - a_{j+1}) - \frac{r}{2} \sum_{j=N+1}^{n-1}j(a_j - a_{j+1}) \\ = -\frac{r}{2}na_n + \sum_{j=1}^{N} S_j(a_j - a_{j+1}) - \frac{r}{2} \sum_{j=N+1}^{n-1}(ja_j - (j+1)a_{j+1}) - \frac{r}{2}\sum_{j=N+1}^{n-1}a_{j+1} \\ = \sum_{j=1}^{N} S_j(a_j - a_{j+1}) - \frac{r}{2}(N+1)a_{N+1} - \frac{r}{2}\sum_{j=N+1}^{n-1}a_{j+1}$$.
Since $\sum a_n$ diverges, the RHS of (*) diverges to $-\infty$ as $n \to \infty$ which contradicts the convergence of $\sum \epsilon_n a_n$.
Thus, $\limsup (\epsilon_1 + \ldots + \epsilon_n)/n \geqslant 0$.  
We use the fact that $a_n$ decreases monotonically but the additional hypothesis that $a_n$ converges to $0$ is not needed.
A: This answer just formalizes my hint from an hour ago, about one particular counter-example: 
Define successive frames that double in size, i.e., frame $k \in \{1, 2, 3, ...\}$ has size $2^k$. 
Define binary $\epsilon_i$ that alternate over successive frames.  That is, $$\{1, 1, -1, -1, -1, -1,1, 1, 1, 1, 1, 1, 1, 1...\}$$ Conclude that the running average of the $\epsilon_i$ values has a nonzero limsup. 
It remains to define a nonnegative and nonincreasing $a_i$ sequence with a sum that diverges but such that $\sum_i a_i \epsilon_i$ converges. 

This is an edit an hour later (requested so someone can undo a downvote). 
