TITLE: Show that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\geq 0$？
QUESTION [11 upvotes]: 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.

REPLY [3 votes]: 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.