
#1
Feb2512, 12:15 PM

P: 9

I have been working on a problem proposed in a math journal, and there is only one thing I need to figure out. Here it is:
Let [itex](a_n)[/itex] be a sequence defined by [itex]a_1 = a[/itex] and [itex]a_{n+1} = 2^n\sqrt{2^n(2^na_n)}[/itex] for all [itex]0 \leq a \leq 2[/itex] and [itex]n \geq 1[/itex]. Find [itex]\lim_{n \rightarrow \infty} 2^n a_n[/itex] in terms of [itex]a[/itex]. What I figured out so far:
Spoiler
Let [itex]A = \lim_{n \rightarrow \infty} 2^n a_n[/itex]. When [itex]a = 0[/itex], [itex]A = 0[/itex]. When [itex]a = \frac{1}{2}[/itex], [itex]A = \frac{\pi^2}{9}[/itex]. When [itex]a = 1[/itex], [itex]A = \frac{\pi^2}{4}[/itex]. When [itex]a = \frac{3}{2}[/itex], [itex]A = \frac{4\pi^2}{9}[/itex]. When [itex]a = 2[/itex], [itex]A = \pi^2[/itex]. I'm still trying to figure it out. Any insight on recurrence equations or limits would be greatly appreciated! Thanks! 



#2
Feb2512, 01:27 PM

P: 192

Here's a tip: if the recurrence is [itex]a_{n+1}=f(a_n)[/itex] where f is continuous, then the limit is a solution to a=f(a). Intuitively, this is because since points near the limit change little, we reason that the limit ought to remain fixed under the recurrence.
Your recurrence is not in that form (since f depends on n), but some trickery may be used to put it there, for instance by considering f(a,n) as a function of a, for fixed, large n. (It'll be tough to justify rigorously, but it may be a good starting point.) 


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