Convergence of $\sum_k \epsilon_k a_k$ implies $(\epsilon_1 + \cdots +\epsilon_k)a_k \rightarrow 0$ Let $\epsilon_k \in \{\pm 1\}$, and suppose $\sum_k \epsilon_k a_k$ is convergent, where $a_k \geq a_{k+1} \geq 0$. Prove that $(\epsilon_1 + \cdots + \epsilon_k)a_k \rightarrow 0$ as $k \rightarrow \infty$.
 A: Let $S_n = \sum_{k=1}^n \epsilon_k$.  For any $\delta > 0$ there exists $N$ such that if $m > n \geqslant N$ we have
$$\left|\sum_{k = n+1}^m \epsilon_k a_k \right| = \left|\sum_{k = n+1}^m (S_k - S_{k-1})a_k \right| < \delta.$$
Suppose $S_n = 0$ infinitely often (and not eventually of the same sign). WLOG we can choose $N$ such that $S_N = 0.$
We have 
$$\delta > \left|\sum_{k = n+1}^m S_ka_k- \sum_{k=n+1}^{m}S_{k-1}a_{k}  \right| \\ = \left|\sum_{k = n+1}^m S_ka_k- \sum_{k=n}^{m-1}S_{k}a_{k+1}  \right| \\ =  \left|S_ma_m - S_na_{n+1} +  \sum_{k=n+1}^{m-1}S_{k} (a_k -a_{k+1})  \right| \\ = \left|(S_m - S_n)a_m +  \sum_{k=n+1}^{m-1}(S_{k}-S_n) (a_k -a_{k+1})  \right| $$
Pick $n$ to be the largest integer with $N \leqslant n < m$ such that $S_n = 0$.  Then all factors $S_k - S_n = S_k$ have the same sign for $n+1 \leqslant k < m$ and the terms within the absolute value are of the same sign (since $a_k \geqslant a _{k+1}$). Hence,  for every $m > N$ we have $|S_m a_m| < \delta,$ and $S_m a_m \to 0$ as $m \to \infty$.
Suppose, on the other hand, that $S_n$ is eventually non-negative.  Again, for any $\delta > 0$ we can find $N$ such that if $N \leqslant n < m \,$ we have the estimate
$$ \left|(S_m - S_n)a_m +  \sum_{k=n+1}^{m-1}(S_{k}-S_n) (a_k -a_{k+1})  \right| < \delta/2,$$
and where $S_k \geqslant 0$ for all $k \geqslant N$. Choose $n$ such that $S_n = \min(S_{N+1}, \ldots S_m).$ Then $S_k - S_n \geqslant 0$ and 
$$|S_m a_m  - S_n a_m| < \delta/2.$$
Hence,
$$|S_m a_m| = S_ma_m < \delta/2 + S_n a_m$$  
Since $a_m \to 0$, for sufficiently large $m$ we have $S_n a_m < \delta/2$ and $|S_m a_m | < \delta$.
A similar argument applies if $S_n$ is eventually non-positive. 
