You can just bound an^2 by 2(an^2)/(1+an) (Using that an is between -1 and 1) to check the part you didnt know how to formalize in problem 1. Great work btw!
Actually, I think your solution to Problem 1 is perfectly valid. Define two partial sums s_N := \sum_{n \le N} a_n and s'_N := \sum_{n \le N} a_n/(1 + a_n). Then, the fact that s_N and s'_N both converges as N \to \infty tells you that s_N - s'_N is also convergent. But s_N - s'_N (as a finite sum) is just the term-wise difference of s_N and s'_N.
In the first problem, for your question @3.45 my argument would be, if we done alpha_n and beta_n denote the nth partial sums of the series, then alpha_n - beta_n = n partial sum of the difference. Since the sum is finite the difference makes sense. Now letting n to infinity we have it.
In problem 1: The termwise subtraction is valid by properties of limits for sequences and the other step can be justified using the limit comparison test
#2 is way easier, and your proof is perfect. I think #1 fails for the reason you point out, that the series are not assumed to converge absolutely. Hence rearranging the terms of alpba and beta not only can be made to diverge but also, with a suitable arrangement, can be made to converge to any arbitrary real number (by the Riemann series theorem or rearrangement theorem). Sadly, I cannot seem to find the solution to #1, but this one doesn't quite work for the reason you mention. I feel like the rest of the proof is basically correct but there's also something about a_n^2 and a_n^4 being nonnegative and thus we can bound them below by 0. There is some trick about how we can multiply the lim sups or lim infs together and show a_n^2 converges then the result immediately follows by the Comparison Test by the fact that 0 < a_n^4 < a_n^2. I am kind of annoyed TBH that I can't quite see it. I'll watch #3 to try to distract myself...by the way your videos are awesome man!! Please keep going. EDIT--nice proof to #3!
As,a physics grad this looks very painful and fun at the same time.
You can just bound an^2 by 2(an^2)/(1+an) (Using that an is between -1 and 1) to check the part you didnt know how to formalize in problem 1. Great work btw!
Actually, I think your solution to Problem 1 is perfectly valid. Define two partial sums s_N := \sum_{n \le N} a_n and s'_N := \sum_{n \le N} a_n/(1 + a_n). Then, the fact that s_N and s'_N both converges as N \to \infty tells you that s_N - s'_N is also convergent. But s_N - s'_N (as a finite sum) is just the term-wise difference of s_N and s'_N.
In the first problem, for your question @3.45 my argument would be, if we done alpha_n and beta_n denote the nth partial sums of the series, then alpha_n - beta_n = n partial sum of the difference. Since the sum is finite the difference makes sense. Now letting n to infinity we have it.
3:45 ?
@@NoNameAtAll2 I mean, the time stamp at which he asked the question in the video.
@@gdineshnathan yeah, youtube uses colons for timestamps
In problem 1: The termwise subtraction is valid by properties of limits for sequences and the other step can be justified using the limit comparison test
I don't think the limit comparison test gives you anything here
#2 is way easier, and your proof is perfect. I think #1 fails for the reason you point out, that the series are not assumed to converge absolutely. Hence rearranging the terms of alpba and beta not only can be made to diverge but also, with a suitable arrangement, can be made to converge to any arbitrary real number (by the Riemann series theorem or rearrangement theorem). Sadly, I cannot seem to find the solution to #1, but this one doesn't quite work for the reason you mention. I feel like the rest of the proof is basically correct but there's also something about a_n^2 and a_n^4 being nonnegative and thus we can bound them below by 0. There is some trick about how we can multiply the lim sups or lim infs together and show a_n^2 converges then the result immediately follows by the Comparison Test by the fact that 0 < a_n^4 < a_n^2. I am kind of annoyed TBH that I can't quite see it. I'll watch #3 to try to distract myself...by the way your videos are awesome man!! Please keep going.
EDIT--nice proof to #3!
Any tips for studying differential equations?
Get a textbook with solutions in the back and do a ton of problems.
@@PhDVlog777 Thank you!
I hate math even more now
From which university are you doing PhD and in what exact domain of mathematics?
He has already told all those things in one of his previous video.