When and for what values of x does lim anxn exist?

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Homework Help Overview

The discussion revolves around the conditions under which the limit of the sequence \( \lim a_n x^n \) exists, focusing on the variable \( x \) and the sequence \( a_n \). Participants explore the implications of different forms of the sequence and the behavior of \( x \) in relation to convergence.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants consider the existence of the limit based on the properties of the sequence \( a_n \) and the variable \( x \). There are attempts to establish inequalities involving \( x \) and subsequences of \( a_n \). Questions arise regarding the behavior of \( x \) under various conditions, such as when \( a_n \) is zero or approaches a constant.

Discussion Status

The discussion is active, with participants providing insights into how the limit's existence may depend on the nature of the sequence \( a_n \). Some guidance has been offered regarding specific cases, such as when \( a_n \) is zero or approaches a constant, but no consensus has been reached on a general solution.

Contextual Notes

Participants note that there are no constraints on the sequence \( a_n \), which adds complexity to the problem. The exploration includes considerations of the limit's behavior based on the magnitude of \( x \) and the relationship between terms in the sequence.

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Homework Statement
Let an be a sequence of real numbers. For what values of x does lim anxn exist?

The attempt at a solution
Let us suppose that lim anxn exist and is equal to b. What can we say about x? Hmm...there is a monotonic subsequence that converges to b, say [itex]a_{k_n}x^{k_n}[/itex]. If this is an increasing sequence, we have that

[tex] a_{k_n}x^{k_n} \le a_{k_{n+1}}x^{k_{n+1}}[/tex]

or equivalently

[tex] \frac{a_{k_n}}{a_{k_{n+1}}} \le x^{k_{n+1} - k_n}[/tex]

Unfortunately I don't get an inequality in terms of x alone. How do I proceed from here? Perhaps I need a further assumption, like [itex]k_{n+1} - k_n = 1[/itex] for all n?
 
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its going to depend on your a_n... if a_n =0 for all n, it exists for every x
 
sorry, actually I misread sequence

if the limit exists, and is b say, then for any e>0 you can choose N such that for all n>N
[tex]|a_nx^n -b|<e[/tex]

you could then think about the beahviour of x when x>|1| and otherwise

though as mentioned previously it will depend on the sequnce [itex]a_n[/itex] , is it for any [itex]a_n[/itex] or are there any contraints on the [itex]a_n[/itex] ?
 
There are no constraints on an. I do know of a couple that might shed some light on this general situation. As you previously mentioned, if the an are all 0 (or are eventually all 0), then the limit exists for all n. If the an are eventually all some nonzero constant, then we know the limit exists for all |x| < 1. Now I believe that if c = lim |an/an+1|, then the limit exists for all |x| < c. This is something I have yet to prove though.
 

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