1. ## Convergent sequence

Note: This is a homework problem.

Suppose that {$\displaystyle {a_n}$} is a bounded sequence of reals such that for any n, $\displaystyle a_n \le \frac{a_{n-1} + a_{n+1}}{2}$.

Show that $\displaystyle b_n = a_{n+1} - a_n$ is an increasing sequence and, $\displaystyle {a_n}$ converges.

I've already done the first part, that $\displaystyle {b_n}$ is an increasing sequence. For the second part, my attempt so far has been that looking at $\displaystyle a_2 \le \frac{a_1 + a_3}{2}$, there are two possibilities:

Case 1: It's possible that $\displaystyle a_3 \ge a_2$. In that case, if for some integer k, $\displaystyle a_{k+1} \ge a_k$, then we can use that $\displaystyle a_{k+1} \le \frac{a_k + a_{k+2}}{2}$ to show that $\displaystyle 2*a_{k+1} \le a_k + a_{k+2} \le a_{k+1} + a_{k+2}$, and so $\displaystyle a_{k+1} \le a_{k+2}$, establishing by induction that {$\displaystyle a_n$} is also an increasing bounded sequence and then convergent.

Case 2: It's also possible that $\displaystyle a_1 \ge a_2$. This is where I'm having trouble. The only way I can see to continue with this is to show that the sequence {$\displaystyle a_n$} is decreasing in this case. After several failed attempts, I'm starting to become skeptical that it actually is decreasing. Is this not going to work, and should I abandon this attempt? If so, does anyone have a suggestion for what direction to go in?

2. In the second case, I assume that $\displaystyle a_k \ge a_{k+1}$ for some positive k. Then, $\displaystyle a_{k+1} \le \frac{a_{k+2} + a_k}{2}$, implying that $\displaystyle 2a_{k+1} \le a_{k+2} + a_k$.

I have no idea where to go after this point. It doesn't seem like I can finish this in this way by showing that the sequence is either monotonically increasing (case 1) or decreasing (case 2). Is there a better way to attempt this?

Thanks for any help!

3. Can you show that $\displaystyle (b_n)$ is bounded above?
Increasing bounded sequences converge.

4. Showing that $\displaystyle (b_n)$ is bounded above follows from the fact that $\displaystyle (a_n)$ is bounded, right? I can just pick an N that's larger than $\displaystyle 2*a_n$ for any n, and then $\displaystyle N > b_n = a_{n+1} - a_n$? So I can conclude that $\displaystyle (b_n} \rightarrow b$, for some b. Then I'd use this to show that the $\displaystyle (a_n)$ sequence converges by taking some $\displaystyle \epsilon > 0$ and noting that for some N, if n > N then $\displaystyle |b + a_{n+1} - a_n| = |b - b_n| < \epsilon$.

I don't really see how the left hand side of that last equality can be simplified to show the convergence of the $\displaystyle (a_n)$ sequence though :/

Edit: Sorry, I'm not trying to fish for answers. I had tried looking at the limit of the $\displaystyle (b_n)$ sequence already and didn't have any luck simplifying the expression. I'm really frustrated trying to see how to evaluate limits in this way, so even a nudge in the right direction would be greatly appreciated.

5. b_n converges to 0 and a_n is a decreasing sequence. when a_n reaches its bound, b_n maxes out at 0