Does the sequence converge, and to what? We have a sequence $\{a_n\}$ 
$$a_0 = 0$$
$$a_{n+1} = \frac{a_{n}}{2} + 1$$
Does it converge? And to what? 

Claim: For $n \geq 0$, $$a_{n}=\frac{2^{n}-1}{2^{n-1}}$$ We can show this inductively. 
Begin with $n=0$ we have $$a_{0}=0=\frac{2^0-1}{2^{-1}} \ \ \checkmark$$ Now let's suppose the result holds for all $k$ where $0<k\leq n$. By definition we have $a_{k+1}=\frac{a_{k}}{2}+1$, and by our induction hypothesis we have $a_{k}=\frac{2^{k}-1}{2^{k-1}}$ Now plug $a_{k}$ into our expression for $a_{k+1}$ and we will get $$a_{k+1}=\frac{\left(\frac{2^{k}-1}{2^{k-1}}\right)}{2}+1$$ $$=\frac{2^{k}-1}{2^{k}}+1$$ $$=\frac{2^{k}-1}{2^{k}}+\frac{2^{k}}{2^{k}}$$ $$=\frac{2\cdot2^{k}-1}{2^{k}}$$ $$=\frac{2^{k+1}-1}{2^{k}}$$ We now know this relationship holds for all $k \in \mathbb{N}$.You can do a little algebra and rewrite this as $a_{n}=2-\frac{1}{2^n}$. Taking the limit of this expression should make it clear that $lim_{n \rightarrow \infty}(2-\frac{1}{2^n})=2$
