# Limit of a sequence - nice one

• Nov 28th 2009, 03:50 AM
Limit of a sequence - nice one
Well this one looks very interesting, it looks pretty obvious but I can't find a way to prove it formally:

Let a,b,c be positive numbers.
Prove:
$lim(a^n+b^n+c^n)^\frac{1}{n} = max(a,b,c)$
• Nov 28th 2009, 04:16 AM
Bacterius
Try to define mathematically what $max(a, b, c)$ is (I believe it is the greatest of these numbers ?). You can't just leave a verbal operation in your limit, you must express it in algebraïc operations. It might be easier (Wondering)
• Nov 28th 2009, 04:44 AM
Plato
Quote:

Let a,b,c be positive numbers.
Prove: $lim(a^n+b^n+c^n)^\frac{1}{n} = max(a,b,c)$

Here is the basic idea.
$\max \left\{ {a,b,c} \right\} = M = \sqrt[n]{{M^n }} \leqslant \sqrt[n]{{a^n + b^n + c^n }} \leqslant M\sqrt[n]{3}$
• Nov 28th 2009, 05:40 AM
chisigma
Let's suppose that $a>b$ and $a>c$ [othervise you can always swap the variables (Nod) ... ]. Because is...

$\sqrt[n]{a^{n} + b^{n} + c^{n}} = a \sqrt{1+(\frac{b}{a})^{n} + (\frac{c}{a})^{n}}$ (1)

... and...

$\frac{b}{a}<1$ , $\frac{c}{a}<1$ (2)

... from (1) and (2) we derive...

$\lim_{n \rightarrow \infty} \sqrt[n]{a^{n} + b^{n} + c^{n}} = a$ (3)

Kind regards

$\chi$ $\sigma$
• Nov 28th 2009, 05:43 AM
Quote:

Originally Posted by chisigma
Let's suppose that $a>b$ and $a>c$ [othervise you can always swap the variables (Nod) ... ]. Because is...

$\sqrt[n]{a^{n} + b^{n} + c^{n}} = a \sqrt{1+(\frac{b}{a})^{n} + (\frac{c}{a})^{n}}$ (1)

... and...

$\frac{b}{a}<1$ , $\frac{c}{a}<1$ (2)

... from (1) and (2) we derive...

$\lim_{n \rightarrow \infty} \sqrt[n]{a^{n} + b^{n} + c^{n}} = a$ (3)

Kind regards

$\chi$ $\sigma$

very nice :) :)!!
I like such proofs!