# Prove $b_n$ is increasing and bounded above given $a_n$ is increasing and bounded above.

If $\{a_n\}$, $n \ge 1$, is increasing and bounded above, then so is $\{b_n\}$, $n \ge 1$, where $b_n = \frac{a_1+a_2+...+a_n}{n}$.

So far, I understand that for a sequence to be increasing, $b_{n+1} \ge b_n$, but every time I try to algebraically manipulate this statement, I can't seem to prove that to be the case.

An attempt:

$$\frac{a_1+a_2+...+a_n+a_{n+1}}{n+1} \ge \frac{a_1+a_2+...+a_n}{n}$$ $$\frac{n}{a_1+a_2+...+a_n} * \frac{a_1+a_2+...+a_n+a_{n+1}}{n+1} \ge 1$$ $$\frac{n}{n+1} + \frac{n(a_{n+1})}{(n+1)(a_1+a_2+...+a_n)} \ge \frac{n+1}{n+1}$$ $$\frac{n(a_{n+1})}{(n+1)(a_1+a_2+...+a_n)} \ge \frac{1}{n+1}$$ $$a_{n+1} \ge \frac{a_1+a_2+...+a_n}{n}$$

As for proving an upper bound exists, I am pretty stumped on how I could prove that for this situation.

• Try proving $b_n \leqslant a_n$ for all $n$. – Daniel Fischer Sep 6 '16 at 13:41
• It's probably helpful to think of $b_n$ as the average of $a_{1},...,a_{n}$. $a_{n}$ is the largest of $a_{1},...,a_{n}$ and the average value cannot be larger than the largest value being averaged. You can also think about what happens to the average of a bunch of values when a new value (larger than the previous average) is added to it. – TravisJ Sep 6 '16 at 13:48

Your attempt is very close to a conclusion. Because $a_i$ is increasing, what can you say about $\frac{a_1+\cdots+a_n}{n}$ as opposed to $\frac{a_n+\cdots +a_n}{n}$? Simplify that last fraction, and finally compare to $a_{n+1}$.

For the boundedness, showing $b_n\leq a_n$ is indeed a good way to go, because $a_n$ is bounded, so then $b_n$ must be too.

• Apologies about the prior comment, I pressed enter assuming it would do a line break! By the time I edited it, 5 minutes had elapsed. Here's the correct comment: "Okay, I think I understand! So, $\frac{a_n+...+a_n}{n}$ simplifies to $a_n$, which I can replace into my previous work as a middle piece between $a_{n+1}$ and $\frac{a_1+...+a_n}{n}$. Because $a_{n+1} \ge a_n$, so too is $a_{n+1} \ge \frac{a_1+...+a_n}{n}$. Similarly, $b_n \le a_n$ can be worked with to get to $a_1+...+a_n \le na_n$ which is true from what you mentioned before, proving $b_n$ is bounded from above." – user366669 Sep 6 '16 at 17:13