Reply to the fresh_42
When I solved the outer sum of question 89, I didn’t use the limit
limn→∞(1−x1)n because I recognized right away that the sum was a geometric series. However, after seeing your approach, I think it would be worthwhile to try starting with the limit next time. If I started with that limit first, I probably wouldn’t have thought to consider the absolute value. I would approach it like this:
limn→∞(1−x1)n=(1−x1)∞, and then I’d conclude that this limit doesn’t go to infinity when
(1−x1)≤1. Since x = 1 is not in the domain, I’d exclude it, leading me to
x<0 but I will miss
x>2 because I didn't consider the absolute value. So yes, I agree that starting with the limit method could partially provide a strong foundation for solving the question.
Reply to the Doctor
After thinking carefully about what you said, I believe I now understand most of the idea. There are two domains to consider: I know the inner domain is
(−1,1), and the outer domain is
(−∞,0)∪(2,∞). If I understand correctly, I need to find the values that belong to both domains, in other words, values that make both the inner and outer series converge. The interval
(2,∞) must be excluded because it causes the inner series to diverge. Similarly,
(0,1) must be excluded because it makes the outer series diverge. That leaves
(−1,0) where both series converge. So essentially, what I did was take the intersection of the two domains, even though I didn’t fully realize it at first, it only became clear to me after working through the process.
Reply to the blamocur
If my explanation to the Doctor was correct, the values of x would be in this domain
(−1,0)