# When does lim a_n x^n exist?

1. Aug 14, 2010

### e(ho0n3

The problem statement, all variables and given/known data
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 $a_{k_n}x^{k_n}$. If this is an increasing sequence, we have that

$$a_{k_n}x^{k_n} \le a_{k_{n+1}}x^{k_{n+1}}$$

or equivalently

$$\frac{a_{k_n}}{a_{k_{n+1}}} \le x^{k_{n+1} - k_n}$$

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 $k_{n+1} - k_n = 1$ for all n?

2. Aug 14, 2010

### lanedance

its going to depend on your a_n... if a_n =0 for all n, it exists for every x

3. Aug 14, 2010

### lanedance

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
$$|a_nx^n -b|<e$$

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

though as mentioned previously it will depend on teh sequnce $a_n$ , is it for any $a_n$ or are there any contraints on the $a_n$ ?

4. Aug 14, 2010

### e(ho0n3

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.