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 :)

  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.