If $\lim_{n \to \infty}2^n|a_{n+1}-a_n|=L>0$ then $a_n$ converges I'm having troubles proving this one.
Since $\lim_{n\to \infty}|a_{n+1}-a_n|=0$ doesn't imply the convergence of $a_n$, I know I need to use the "rate of convergence", namely that the difference between every consecutive elements of $a$ is "about" $2^{-n}$:
$$\frac{L-\varepsilon}{2^n}<|a_{n+1}-a_n|<\frac{L+\varepsilon}{2^n}$$
and then
$$|a_{n+1}|<|a_n|+\frac{L+\varepsilon}{2^n}  $$
so that 
$$|a_{n+1}-a_n|<\left||a_n|+\frac{L+\varepsilon}{2^n} -a_n \right|$$
However, I can't seem to make progress.
Thanks for your help. 
 A: Hint: Use Cauchy's criterion for the convergence of a sequence. Use the fact that $|a_n - a_m| = |a_n - a_{n-1} + a_{n-1} - \ldots - a_m|$ (if $n > m$), the triangle inequality and the information you have about the limit to bound the expression accordingly.
A: You have for an arbitrarily fixed $\epsilon>0$
$$|a_{n+1}-a_n|<\frac{L+\varepsilon}{2^n}$$ for all $n\ge N_\epsilon$. Take $\epsilon=1$ and find $N_1$.
Then chose an arbitrary $\delta>0$ 
$$|a_{n+k}-a_{n}|\leq |a_{n+k}-a_{n+k-1}|+...+|a_{n+1}-a_{n}|\leq (L+1)\sum\limits_{i=0}^{k-1}{\frac{1}{2^{n+i}}},\,\forall n\ge N_1,\,\forall k>0$$ 
Because the series 
$$\sum\limits_{n=0}^{\infty}{\frac{1}{2^n}}$$ converges, its tail goes to $0$ and so 
$$\exists N_\delta: \sum\limits_{i=0}^{\infty}{\frac{1}{2^{n+i}}}\leq \frac{\delta}{L+1},\,\forall n\ge N_\delta$$ 
Finally, for all $n\ge \max{\{N_1,N_\delta\}},\,k>0$ we have 
$$|a_{n+k}-a_{n}|\leq (L+1)\sum\limits_{i=0}^{k-1}{\frac{1}{2^{n+i}}}\leq (L+1)\sum\limits_{i=0}^{\infty}{\frac{1}{2^{n+i}}}\leq \delta$$
which means that $\{a_n\}_{n\in\mathbb N}$ is a Cauchy sequence in $\mathbb R$, which is complete and so the sequence is convergent.
