If $x\_{a+1}$-$x\_{a}$ converges to $0$ and $x\_{2a}$-$2x\_{a}$ converges to $0$ , does that imply $x\_a$ converges to $0$? 



Given a sequence of real numbers,do the following conditions suffice to guarantee convergence to 0? 
Yes. There's probably a clever proof, but here's a non-clever 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 
`\[
x_{2^{k+1}a}-x_{2^ka}\geq x_{2^ka}-\frac12\geq2^{k-1}.
\]`
That is, as $b$ increases from $2^ka$ to $2^{k+1}a$, a distance of $2^ka$, the value of $x\_b$ increases by at least $2^{k-1}$. Therefore, at some step along the way, from some $b$ to $b+1$, the value of $x\_b$ must have increased by at least $1/(2a)$. This happens repeatedly, with the same $a$, because $k$ can increase arbitrarily, and it therefore contradicts the hypothesis that $x\_{b+1}-x\_b$ tends to 0. 


