sequence for n >=1 : (a^n + b^n) ^1/n

[dts5044]

Let a_n = (a[sup:2tjiyd9f]n[/sup:2tjiyd9f] + b[sup:2tjiyd9f]n[/sup:2tjiyd9f])[sup:2tjiyd9f]1/n[/sup:2tjiyd9f], for n >=1, where a, b are two fixed real numbers >= 0. When is the sequence (a_n) convergent, and what is its limit?

 

[tkhunny]

Binomial Theorem Expansion?

First, you may wish to rewrite: \(\displaystyle (a^{n} + b^{n})^{1/n} = a(1 + (b/a)^{n})^{1/n}\)

 

[Dr. Flim-Flam]

Very good ,TKhunny, but where do you go from here?

 

[pka]

Let a_n = (a[sup:16057zhe]n[/sup:16057zhe] + b[sup:16057zhe]n[/sup:16057zhe])[sup:16057zhe]1/n[/sup:16057zhe], for n >=1, where a, b are two fixed real numbers >= 0. When is the sequence (a_n) convergent, and what is its limit?

This is a rather important sequence. The proof and the limit is the same for any finite collection of positive numbers.
If \(\displaystyle m = \max \left\{ {a,b} \right\}\) then
\(\displaystyle m = \sqrt[n]{{m^n }} \leqslant \sqrt[n]{{a^n + b^n }} \leqslant \sqrt[n]{{2m^n }} = m\sqrt[n]{2}\).
Well of course we all know that \(\displaystyle \left( {\sqrt[n]{2}} \right) \to 1\).
So the squeeze is on: \(\displaystyle \left( {\sqrt[n]{{a^n + b^n }}} \right) \to m\).

For a finite collection of positives: \(\displaystyle M = \max \left\{ {a_1 ,a_2 ,a_3 , \cdots ,a_K } \right\} \Rightarrow \left( {\sqrt[n]{{\sum\limits_{j = 1}^K {\left( {a_j } \right)^n } }}} \right) \to M\)

 

[tkhunny]

Very good ,TKhunny, but where do you go from here?

It was just so obvious that it didn't even cross my mind that the solution would not be obvious from the expansion. Still sleeping, I guess.

 

[royhaas]

This is a limiting example of a generalized mean.

 

[pka]

It was just so obvious that it didn't even cross my mind that the solution would not be obvious from the expansion.

It is not at all clear what that means. Nor is it clear that this problem has been addressed outside my posting. Here is a problem related to this topic.
\(\displaystyle \left( {\sqrt[n]{{3^{ - n} + 2^n + 2^{ - n} }}} \right) \to ?\)

 

[daon]

Tkhunny was on the right track, though...?

If a>=b then

\(\displaystyle (a^n+b^n)^{\frac{1}{n}} = a(1 + (b/a)^{n})^{1/n} \rightarrow a\)

as the second factor, \(\displaystyle 1 \le (1+\frac{b^n}{a^n})^{\frac{1}{n}} \le 2^{\frac{1}{n}} \rightarrow 1\).