I'm stuck with the following problem:
Let $(X_k)_{k=1}^\infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^\infty$ such that $$P\left(\lim_{k\rightarrow\infty}\frac{X_k}{c_k}=0\right)=1.$$
With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at $$P\left(\lim_{k\rightarrow\infty}\frac{X_k}{c_k}\neq 0\right)\leq \sum_{n=1}^\infty P\left(\limsup_{k\rightarrow\infty}|X_k|>\frac{c_k}{n}\right)$$ and it would be sufficient to find $(c_k)_{k=1}^\infty$ such that $$\sum_{k=1}^\infty P\left(|X_k|>\frac{c_k}{n}\right)<\infty.$$ If the $X_k$ were identically distributed, I think a construction of the form $c_k=k\cdot q_{1-1/(k^2)}$ could work, where $q_\alpha$ is an $\alpha$-quantile of $|X_1|$, but I can't do this in the general case.
How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?
If $P\{|X_k| >\frac {c_k} {2^{k}}\} <\frac 1 {2^{k}}$ then $\sum_k P\{|X_k| >\frac 1 {2^{k}}c_k\} <\infty$ and Borel Cantelli Lemma tells you that $\frac {X_k} {c_k} \to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P\{|X_k| >t\} \to 0$ as $t \to \infty$].