Help with power series sum_{n=1, infinity} [(x-2)^n]/[(n^2)(3^n)] (Find interval of convergence)

[roineust]

Here is the question from the quiz at Khan academy, the problem is that, what i think is the correct answer, i can't find in any option to choose:

 

[fresh_42]

What is your idea? Which terms are the ones that count?

 

[roineust]

1<x<3 i didn't check the ends 1 and 3 yet, but these are the numbers I'm getting.

 

[fresh_42]

What if you set [imath] x=5 [/imath]?

But guessing numbers doesn't help us. That's why I asked for the dominating parts in the sum.

 

[roineust]

Then 5 is indeed larger than 1, but it is not smaller than 3.

 

[roineust]

Here is my calculation:

 

[fresh_42]

What is required if you look at [imath] (x-2)^n \cdot \left(\dfrac{1}{3}\right)^n \cdot \dfrac{1}{n^2}[/imath]? Which terms grow fastest, and which are irrelevant for large [imath] n [/imath]?

 

[fresh_42]

We need [imath] \dfrac{|x-2|}{3} < 1.[/imath]

 

[roineust]

OK, i see my mistake, its 1/3 and not 1.

 

[roineust]

I should have been more careful and one day also chat GPT will be less of a stubborn hallucinator.

 

[fresh_42]

OK, i see my mistake, its 1/3 and not 1.

The situation would have been more interesting if it wasn't [imath] n^2 [/imath] but [imath] n [/imath] in the denominator. In that case, we would have got a different behavior at the two boundaries of the interval.

 

[roineust]

The situation would have been more interesting if it wasn't [imath] n^2 [/imath] but [imath] n [/imath] in the denominator. In that case, we would have got a different behavior at the two boundaries of the interval.

Yes, if I'm not wrong, then one side is still an alternating series and converges, but the other side becomes a diverging harmonic series.