The discussion revolves around finding the radius of convergence for the series \(\Sigma \frac{nx^{2n}}{2^{n}}\) using the ratio test. Participants are exploring the implications of applying the ratio test and the conditions under which the series converges.
There is an ongoing exploration of the correct interpretation of the results from the ratio test. Some participants suggest different approaches and clarify misunderstandings about the absolute value and the relationship between \(x\) and \(\sqrt{2}\). While there is no explicit consensus, several participants are providing guidance and corrections to earlier statements.
Participants are navigating through various interpretations of the radius and interval of convergence, with some expressing uncertainty about their previous statements and the implications of the ratio test results.
Lamoid said:I think that should be x^2 < 2 actually. Since it will converge if x^2 / 2 is less than 1.Also, I don't believe you need to make it absolute value of x^2 since x^2 will always be positive.
Lamoid said:Remember you have x^2 < 2. Just solve it for x. And about that absolute value thing, disregard what I said.
BoundByAxioms said:So, I get x<\pm\sqrt{2} (I know that doesn't make sense).
Lamoid said:Ok, well my instruction in my first post to remove the absolute value signs was dumb. If you take the root of both sides you get root of x^2 < root of 2. Remember that the root of x^2 is the absolute value of x anyway. So you see why what I told you was such a waste of time?
To clarify, you have:
\left|x^{2}\right|<2 then
\left|x\right|<\sqrt{2}
Haha, ok thank you! I see your an east coaster, lol. For me it's only 12.Lamoid said:This is why I don't stay up to 3 AM. I don't know why I said to take the root. You would only do that if you wanted the interval of convergence. You don't have to take the root. You just want to leave it as absolute x^2 < 2. Thus you say the radius of C is 2.
HallsofIvy said:The instruction to take the root originally was correct. That is because the "2" in the exponent of x.
By the ratio test we must have
\frac{(n+1)|x|^{2(n+1)}}{2^{n+1}}\frac{2^n}{n|x|^n}= \frac{n+1}{n}\cdot\frac{1}{2}|x|^n< 1
The limit of that is not "1/2", it is (1/2)|x|2 which must be less than 1: (1/2)|x|2< 1 so |x|2< 2 and |x|< \sqrt{2}.
HallsofIvy said:More correctly, it will converge absolutely for for (1/2)x2|< 1, diverge for (1/2)x2|> 1 and may converge conditionally or diverge when (1/2)x2|= 1.
That is, of course, because the ratio test says that a sum, \sum_{n=0}^\infty a_n of positive numbers will converge if lim a_{n+1}/a_n< 1 converge for lim a_{n+1}/a_n >1 and may converge or diverge if that limit is 1.
HallsofIvy said:And, by the way, "the radius is -\sqrt{2}< x< \sqrt{2}" is wrong. That is the interval of convergence. The radius of convergence is \sqrt{2}