Existence of a.s . converges of weighted sum of random variables . 
Let $\xi_i$ be random variables, not necessarily independent, show that there exists some $c_i>0$, s.t. the series $S_n = c_1\xi_1+\cdots+c_n\xi_n$ converges a.s..

If $\xi_i$ are bounded by some $M_i$ a.s, then it's easy, but I have no idea how to proceed otherwise, I think I need some hints.
 A: Mike Earnest gives a truncation type answer.  Here are two other ways to do it, one using the Markov inequality as you had suggested in a comment. 
Idea: Define a nonnegative bound on the tail of the sequence: 
$$T_n = \sum_{i=n}^{\infty} |c_i||X_i| \quad, \forall n \in \{1, 2, 3, ...\}$$ If $P[T_n>1/2^n] \leq 1/2^n$ for all $n \in \{1, 2 ,3, ...\}$, then $\{T_n>1/2^n\}$ happens for only finitely many $n$ (with probability 1) and so the tail goes to zero with probability 1.  
1) Markov inequality (assuming $E[|X_i|]<\infty$ for all $i$): For all $n \in \{1, 2, 3, ...\}$ we have: 
\begin{align}
P[T_n> 1/2^n] &\leq \frac{E[T_n]}{1/2^n} \\
&= \underbrace{2^n\sum_{i=n}^{\infty} |c_i|E[|X_i|]}_{\theta_n}
\end{align}
then define $c_i> 0$ such that $c_i E[|X_i|]=d_i$ where $d_i$ is any sequence whose tail decays sufficiently fast so that $\theta_n \leq 1/2^n$. 
2) Union bound (as in my above comment): 
$$ \left\{T_n > 1/2^n\right\} \subseteq \cup_{i=n}^{\infty} \left\{|c_i||X_i| > \frac{1/2^i}{2^{i+1}}\right\} $$
so $$P[T_n > 1/2^n] \leq \sum_{i=n}^{\infty} P\left[|c_i||X_i|> \frac{1/2^i}{2^{i+1}}\right] = \sum_{i=n}^{\infty} \underbrace{P\left[|X_i|>\frac{1}{2|c_i|4^i}\right]}_{\beta_i} $$
then design the $c_i> 0$ for the right-hand-side to be $\leq 1/2^n$.
A: As you noted, this would be easy if each $|\xi_i|\le M_i$. In general, choose some numbers $M_i$, and let $X_i$ be the truncation of the $\xi_i$ at $M_i$:
$$
X_i = \xi_i\cdot 1(|\xi_i|\le M_i)
$$
Now you have that $|X_i|\le M_i$, so you can choose $c_i$ so $\sum c_iX_i$ converges a.s.
What about the $\xi_i$? The idea is to choose the $M_i$ large enough so that 
$$
P(\xi_i\neq X_i\text{ infinitely often})=P(|\xi_i|>M_i\text{ infinitely often})=0
$$
so that with probability one, $\sum c_iX_i$ converges if and only if $\sum c_i\xi_i$ does.
