# Convergent Sequence

• October 17th 2012, 03:07 PM
lovesmath
Convergent Sequence
Suppose a,b>0. Does the sequence {(a^n + b^n)^(1/n)} from n=1 to infinity converge? If so, find the limit.

Not sure where to start...
• October 17th 2012, 03:47 PM
johnsomeone
Re: Convergent Sequence
Factor out the larger of a and b. You'll need to consider the two cases where they're equal and not equal separately.

If the one case isn't clear to you, you could use the identity $p = e^{\ln(p)} \ \forall \ p>0$, and then look at the limit in the exponent.

Question:

$\text{If } \lim_{n \to \infty} q_n \text{ exists and equals } L, \text{ then } \lim_{n \to \infty} e^{q_n} = e^L \text{ is true because }$

$\text{ the function } f(x) = e^x \text{ is ?what? }$
• October 17th 2012, 04:10 PM
Plato
Re: Convergent Sequence
Quote:

Originally Posted by lovesmath
Suppose a,b>0. Does the sequence {(a^n + b^n)^(1/n)} from n=1 to infinity converge? If so, find the limit.

This is such a well known problem.
Suppose that $a>b>0$ then $a = \sqrt[n]{{a^n }} \leqslant \sqrt[n]{{a^n + b^n }} \leqslant a\sqrt[n]{2} \to a$

It can be extended to more than two variables.
• October 17th 2012, 04:23 PM
lovesmath
Re: Convergent Sequence
How did you get the 2 under the radical at the end of the inequality?
• October 17th 2012, 05:45 PM
Plato
Re: Convergent Sequence
Quote:

Originally Posted by lovesmath
How did you get the 2 under the radical at the end of the inequality?

It is simple arithmetic.
Because $b^n\le a^n$ we have $\sqrt[n]{{a^n + b^n }} \leqslant \sqrt[n]{{a^n + a^n }} = \sqrt[n]{{2a^n }} = a\sqrt[n]{2}$