Proof of sequence in inequality by induction

Question

I'm not seeing what's so convincing about the following proof.

Let $x_1 = 1$ and for each $n \in N$, define $x_{n+1} = (\frac{1}{2}) x_n +1$.

To show $x_n \leq x_{n+1}$ for all values of $n \in N$, I'm told that all we need to do is define the following: $\frac{1}{2} x_n +1 \leq \frac{1}{2} x_{n+1} + 1$, which is the desired conclusion of $x_{n+1} \leq x_{n+2}$ .

I'm not seeing how this proof is so compelling. For example, what's preventing me from setting the inequality in the opposite direction, like:

$\frac{1}{2} x_n +1 \geq \frac{1}{2} x_{n+1} + 1$ ?

I've looked at a number of other related posts, but nothing seems to address this issue.

Answer 1

To prove that this sequence is increasing, we first need to show that it is bounded above. Let's prove that for all $n: x_n < 2$. For the first term, this holds by assumption: $x_1 = 1 < 2$. Now, let's assume that $x_n < 2$, but then $x_{n+1} = x_n/2 + 1 < 2/2 + 1 = 2$, which means that by induction, this holds for all $x_n$.

Now, let's prove that the sequence is increasing: Suppose that this is not the case, i.e., $x_n \geq x_{n+1}$ for some $n$. Then $x_n \geq x_n/2 + 1 \Rightarrow 2x_n \geq x_n + 2 \Rightarrow x_n \geq 2$, which is a contradiction, since we have already proven that $x_n < 2$. Thus, for all $n: x_n < x_{n+1}$.

Answer 2

There's nothing preventing us from trying to prove the opposite direction:

$$(1)\ \frac12x_n+1\geq\frac12x_{n+1}+1.$$

It's just that this does not lead to any useful conclusion, so we have to try proving the other direction

$$(2)\ \frac12x_n+1\leq\frac12x_{n+1}+1.$$

After all, if we try to prove $x_n$ is monotone (and not constant), only one of the above cases is true. Therefore, I would not say case $(2)$ is more compelling than case $(1)$, it's simply because $(2)$ turns out to be the correct direction of solving the problem.

We can of course try to prove $(1)$ first, and we will find it does not lead to the solution, we then try $(2)$ instead. So it's just trial and error.

For example, if we have $x_1=3$ instead of $x_1=1$, then $(1)$ will be the correct direction (situation flipped). Hence, the correct choice is all situational, but sometimes by observing the problem we could possibly see which case is likely the correct one and avoid trying the other one.

(For this answer, I am assuming the question is simple enough that $x_n$ is monotone. For more complex questions where $x_n$ is not monotone, both cases fail and we will need more tools to find the solution.)