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? 
 A: Hint: show by induction that the sequence $a_n$ increase. Trick to do this is to see that
$a_{n+1}-a_n=a_n/2-a_{n-1}/2$.
Now this shows that the sequence is increasing. What else do you need to see it converges?
Further, suppose that the limit is $L$. Then $L=L/2+1$, as $\lim a_{n+1}=\lim a_n $
A: HINT:
$$a_{n+1}=\frac{a_n}2+1\iff 2a_{n+1}=a_n+2$$
Let $a_n=b_n+c\implies2(b_{n+1}+c)=b_n+c+2$
Set $c=2$ to find $b_{n+1}=\dfrac{b_n}2$
So, $\lim_{m\to\infty}b_m=\dfrac{b_0}{2^m}=0$ as $b_0=a_0-2=\cdots$
A: In my opinion the best way to solve this problem is as many answers have done: show that the series is increasing and bounded and so on.
However, for this problem it is not so hard to find an explicit formula for the $n$'th term of the series and thereby showing the limit explicitly.
Let $y_n= a_n - a_{n-1}$ then the recurence relation becomes
$$y_{n+1} = \frac{y_n}{2} \to y_{n} = \left(\frac{1}{2}\right)^n$$
and therefore $$a_n =\sum_{k=1}^n (a_{k}-a_{k-1}) = \sum_{k=1}^n y_k = \sum_{k=1}^n\left(\frac{1}{2}\right)^k = 2-\left(\frac{1}{2}\right)^{n-1}$$
From this is follows directly that $\lim_{n\to\infty}a_n = 2$.
A: 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 $lim_{n \rightarrow \infty}(2-\frac{1}{2^n})=2$
A: To see if the sequence converges, you have to find if $a_n$ is monotone and bounded.
Then if $a_n$ is incresing you have to check if the sequence is upper bounded, and if $a_n$ is decreasing you have to check if the sequence is bounded below.
Having shown that the sequence converges, then set $L$ the limit of the sequence.
That means that $a_n \underset{n \rightarrow \infty}{\longrightarrow } L$ and $a_{n+1} \underset{n \rightarrow \infty}{\longrightarrow } L$.
So taking the limit at the relation for $n \rightarrow \infty$, you will get a relation for $L$.
A: $$a_{0}=0\\a_{n+1}=\frac{a_{n}}{2}+1\\a_{1}=\frac{a_{0}}{2}+1=1\\a_{2}=\frac{a_{2}}{2}+1=1+\frac{1}{2}\\a_{3}=\frac{\frac{3}{2}}{2}+1=1+\frac{3}{4}\\a_{4}=\frac{\frac{7}{4}}{2}+1=1+\frac{7}{8}\\a_{5}=\frac{\frac{15}{8}}{2}+1=1+\frac{15}{16}\\a_{n}=\frac{2^{n-1}-1}{2^{n}}+1\\...\\a_{n}<2\\so\\it-converge
$$
