How can I show that this sequence is monotone and bounded?

Question

How can I show that the following sequence is monotone and bounded?

$a_1=1, a_{n+1}= \frac{a_n}{2}+1, \forall n \in \mathbb{N}$

The exercise asks to find the limit of the sequence (if it exists), and suggests to first, show that $(a_n)_n$ is monotone and bounded.

Answer 1

You can show it is Montone and bounded by induction easily. First, check that $a_1\leq 2$ and $1=a_1 < a_2=1.5$. Now suppose that $a_n<a_{n+1}<2$ for some $n$. Then, we have $$a_n/2+1</a_{n+1}/2+1<2/2+1=2$$ which implies that $a_{n+1}<a_{n+2} <2$.

you can generally show that if $f(x)$ is a monotone function the $a_{n+1} =f(a_n)$ is a monotone sequence.

Answer 2

Sure you can. Interpret the sequence as $a_{n+1}=\frac{a_n+2}{2}$, and is therefore the average of the previous term with $2$. You started below $2$, so what happens when you average against it?

Answer 3

Claim: $1\le a_n\le 2$ for all $n$. The base case is given, and if $1\le a_n\le 2$, then $\frac12\le \frac{a_n}{2}\le 1$, so $\frac32\le\frac{a_n}{2}+1\le 2$.

Answer 4

The first thing to do is determine a candidate value for the limit. If the limit exists, then we have

$$\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} a_{n + 1} = \lim\limits_{n \to \infty} \frac{a_n}{2} + 1 = 1 + \frac{\lim\limits_{n \to \infty} a_n}{2}$$

Let $L = \lim\limits_{n \to \infty} a_n$, if it exists. Then $L = 1 + \frac{L}{2}$, so $L = 2$.

So if the limit exists, the limit is 2.

Now, define $b_n = 2 - a_n$. We would expect to see that $\lim\limits_{n \to \infty} b_n = 0$.

Then we see that $b_0 = 1$ and $b_{n + 1} = 2 - a_{n + 1} = 2 - (1 + \frac{a_n}{2}) = \frac{2 - a_n}{2} = \frac{b_n}{2}$.

Therefore, we see easily that $b_n = \frac{1}{2^n}$ for all $n$. Therefore, $\lim\limits_{n \to \infty} b_n = 0$, and thus $\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} 2 - b_n = 2 - \lim\limits_{n \to \infty} b_n = 2$.

Alternately, you can note that since $b_{n + 1} = \frac{b_n}{2}$, that $b_n$ is always positive and that $b$ is decreasing. Therefore, $a$ is increasing, and $a_n$ is always less than $2$. So $a$ must converge, and thus must converge to 2.