# Monotonically decreasing sequence $(a_n) \to$ mo. dec. sequ. $(a_1+a_2+\cdots+a_n)/n$

I think this isn't quite difficult, however I don't get the point..

I have to prove:

$(a_n)$ is a monotonically decreasing sequence. Show, that the sequence $\frac{a_1 + a_2 + \cdots + a_n}{n}$ is monotonically decreasing, too

I thought about using something like Cauchy-Limit or so.. Since the sequence $\frac{a_1 + a_2 + \cdots + a_n}{n} = \frac{1}{n} (a_1 + a_2 + \cdots+ a_n)$ looks quite interesting so far..

If you have any hints for me, I'd be glad to use them :)

• Notice that by monotonicity we have $a_1+a_2+\ldots+a_n>na_{n+1}$. It follows that $n(a_1+a_2+\ldots+a_n)+(a_1+a_2+\ldots+a_n)>n(a_1+a_2+\ldots+a_{n+1})$. See if you can finish it from here. – Jared May 21 '13 at 9:04

In a sense, the proof writes itself. The details depend on whether we use $\lt$ or $\le$ in the definition of monotonically decreasing. Let's use $\le$. We want to know whether $$\frac{a_1+a_2 +\cdots+a_{n+1}}{n+1} \overset{?}{\le} \frac{a_1+a_2+\cdots+a_n}{n}.\tag{1}$$ Algebraic manipulation shows that this is equivalent to $$na_{n+1}\overset{?}{\le} a_1+a_2+\cdots+a_n.\tag{2}$$ Since $a_{n+1}\le a_i$ for $i=1,2,\dots, n$, Inequality $(2)$ is obvious.

• Oh, well.. Thank you :) very much :) – Vazrael May 21 '13 at 10:10
• You are welcome. Sorry, I missed the hint request part of your post. – André Nicolas May 21 '13 at 10:13