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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" sub-sequences. It is a non-void closed set by construction. The property $x_{a+1}-x_a\rightarrow0$ tells us that $\Omega$ is a connected set. The property $x_{2a}-x_a\rightarrow0$ 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$\ell/2\in\Omega$. 1 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" sub-sequences. It is a non-void closed set by construction. The property$x_{a+1}-x_a\rightarrow0$tells us that$\Omega$is a connected set. The property$x_{2a}-x_a\rightarrow0$tells us that$2\Omega=\Omega$. Therefore$\Omega\$ can only be equal to one of the four sets $$\{0\},\quad[0,+\infty],\quad[-\infty,0],\quad{\mathbb R}.$$