Let $\displaystyle ( a_{n})$ be a bounded sequence that satisfies $\displaystyle a_{n+1} =a_{n} -\frac{1}{2^{n}}$.

Let $ $$\displaystyle \lim \sup \ ( a_{n}) =a^{*} \land \ \liminf \ ( a_{n}) =a_{*}$

Let $\displaystyle m\in \mathbb{N}$ be arbitrary and let $\displaystyle n >m$, then we have

$\displaystyle a_{m+1} \geq a_{m} -\frac{1}{2^{m}}$

$\displaystyle a_{m+2} \geq a_{m+1} -\frac{1}{2^{m+1}}$

...

...

...

$\displaystyle a_{n+1} \geq a_{n} -\frac{1}{2^{n}}$

Adding all these to get: $\displaystyle a_{n+1} -a_{m} \geq \frac{1}{2^{n}} -\frac{1}{2^{m-1}} \Longrightarrow \forall \epsi…