Limit of infinite sequence raised to a real number power

issisoccer10
Messages
35
Reaction score
0

Homework Statement



Let {an} be a sequence with positive terms such that lim an = L > 0.

Prove lim (an)x = Lx.

Homework Equations



If x is a real number, there exists an increasing rational sequence {rn} with limit x.

A monotone sequence {an} is convergent if and only if {an} is bounded.

ax = lim arn.

The Attempt at a Solution



I know that each of the individual elements of the sequence {(an)x} converge to (an)x, but I do not know what else I can do from here.

Any help would be greatly appreciated. Thanks!
 
Physics news on Phys.org
It may help to know that every convergent sequence in \mathBB{R} contains a monotone subsequence. The power function f(x) = x^a preserves monotonicity (although it may reverse it for negative powers). What would be a bound for such a sequence? Why would it have to be the least upper bound/greatest lower bound? What does that say about the rest of your sequence?
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top