If $x_{a+1}$$x_{a}$ converges to $0$ and $x_{2a}$$2x_{a}$ converges to $0$ , does that imply $x_a$ converges to $0$?
Yes. There's probably a clever proof, but here's a nonclever one. Suppose you had a sequence satisfying your hypotheses but not converging to 0. Multiplying it by a suitable constant (which doesn't affect the hypotheses), you can assume that $x_a>1$ for infinitely many $a$, in particular for some $a$ so large that $x_{2b}2x_b<\frac12$ for all $b\geq a$. Applying that inequality repeatedly, with $a,2a,4a,\dots$ as $b$, you get that $x_{2^ka}>(2^k+1)/2$ for all integers $k\geq0$. Then



Just a very quick argument which reduces the possibilities: Let $\Omega\subset{\mathbb R}\cup\{\pm\infty\}$ be the $\omega$limit set of the sequence, that is the set of limits of "converging" subsequences. It is a nonvoid closed set by construction. The property $x_{a+1}x_a\rightarrow0$ tells us that $\Omega$ is a connected set. The property $x_{2a}2x_a\rightarrow0$ tells us that $2\Omega=\Omega$. Therefore $\Omega$ can only be equal to one of the four sets $$\{0\},\quad[0,+\infty],\quad[\infty,0],\quad{\mathbb R}.$$ Edit. It was commented that the second property gives only an inclusion, of $2\Omega$ into $\Omega$. Actually, it does give also the reverse inclusion (hence the equality), when combined with the first property: Let $\ell$ be the limit of some subsequence $x_{n_k}$. Because of the first property, we may suppose that $n_k=2m_k$ is even. Then $\ell/2$ is the limit of $x_{m_k}$, hence $\ell/2\in\Omega$. 

