What is the domain of the A(x) function?

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The discussion focuses on determining the domain of the function A(x) defined by an infinite series. It is established that the series converges for |x| ≤ 1 based on the comparison test with the convergent series x^(3n). However, there is uncertainty about whether values of |x| > 1 might also lead to convergence due to the diminishing effect of the denominator. The ratio test is suggested as a method to analyze convergence, indicating that the series converges for all values of x. Ultimately, the conclusion remains that while convergence is confirmed for |x| ≤ 1, further proof is needed to explore the behavior for |x| > 1.
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This is my first encounter with this function. In the question at hand it's defined as

A(x) = 1 + x^3/(2)(3) + x^6/(2)(3)(5)(6) + x^9/(2)(3)(5)(6)(8)(9) + ...
and on like that to infinity. The question is to find the domain of this function. Here's what I have so far. The domain would mean the values for which the sum converges. Since this function is smaller than the summation of x^(3n), from n=o to n=infinity, and that function is convergent for |x|<=1, this function must also be convergent for |x|<=1 by the comparison test (a sort of modified squeeze theorem). But this alone doesn't prove that values of |x|>= 1 won't work, it simply shows that values less than 1 will work. Are there values greater than 1 that still cause the function to converge due the the decreasing effect of the denominator? If so, or if not, how can it be proven? I suspect that values >=1 won't work but i don't know how to prove it. Can anyone help me out?

Sorry that I don't know how to make the equations look nice.
 
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Hey so if you write the series as:

x^3n/2*3...((3N-1)(3N))

Then you can take a ratio test pretty easily

The ratio test gives you

lim x/(3N+2)(3N+3)
n->Inf.

Which means that for all values of x, the series will converge.
 
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