Are there any convergence tests that are extensions/generalizations of the root test? The ratio test has a bunch of extensions (Raabe's test, Gauss's test, Kummer's test, etc.).
Thank you! Clean explanation! However, I have a doubt: you proved the condition 2 in terms of lim sup. If we follow the strategy what you followed in condition 1 (using geometric series), we can find condition 2 in terms of lim inf. In that case condition2 can be translated as: if lim inf > 1, the series is divergent. (Because now we can find a geometric series with common ratio greater than 1. And the sum of our series is greater than the sum of this geometric series.) And since lim inf 1, this is particularly your condition 2. Is this reasoning correct? Please help. Also, I have seen in most resources the condition of divergence is given in terms of lim sup like you gave. But the condition what I proved above (in terms of lim inf) is stronger (because this implies your condition 2).
Regarding the main idea, how do we know that the sequence |an|^1/n doesn't approach the limsup (i.e alpha) from above? We're saying that for large N it's less than or equal to alpha but i'm not sure why that is true
|a_n|^{1/n} may be greater than \alpha for finite numbers of n but not for infinite numbers of n. If for infinite number of values of n, |a_n|^{1/n} > \alpha, then lim sup would be greater than \alpha (because in that case there exists k such that the supremum of k-tail of the sequence would be greater than \alpha, contradiction).
Weird proof of the last claim: If a series converges, but does not converge absolutely, alpha cannot be greater than 1 or less than 1 (by the first two claims), so it must be equal to 1. Since such series exist, it must be possible for a series with alpha equal to 1 to converge. And since it doesn't converge absolutely, and taking absolute values term by term obviously doesn't affect alpha, it is possible for a series with alpha equal to 1 to diverge. Proof by two examples I didn't bother to identify? Your example additionally shows that a series with alpha equal to 1 could converge absolutely, though.
That series converge to a rational number. One question, does the slowly convergent series 1/(k*log(k)^2) converge to a transcendental? Every term is transcendental, so does the infinite sum converge to a transcendental?
@@drpeyam It converges to approximately 2.1097, but how can we prove or disrpove that this number is transcendental. I forgot to say that in the sum, k goes from 2 to infinity.
Aditya Dwivedi If e + pi is algebraic, that means that the algebraics isn’t closed under addition. But the naturals, the integers, the rationals, even the Reals as a whole, are all closed under this operation. How does this work?
What a mad coincidence. I just got off my Calc BC class from learning about the Root Test.
Are there any convergence tests that are extensions/generalizations of the root test? The ratio test has a bunch of extensions (Raabe's test, Gauss's test, Kummer's test, etc.).
Thank you! Clean explanation! However, I have a doubt: you proved the condition 2 in terms of lim sup. If we follow the strategy what you followed in condition 1 (using geometric series), we can find condition 2 in terms of lim inf.
In that case condition2 can be translated as: if lim inf > 1, the series is divergent. (Because now we can find a geometric series with common ratio greater than 1. And the sum of our series is greater than the sum of this geometric series.) And since lim inf 1, this is particularly your condition 2. Is this reasoning correct? Please help.
Also, I have seen in most resources the condition of divergence is given in terms of lim sup like you gave. But the condition what I proved above (in terms of lim inf) is stronger (because this implies your condition 2).
omg I just need it, like literally right now
Which book's section 11 are you referring ? @ 17:17
Ross
Regarding the main idea, how do we know that the sequence |an|^1/n doesn't approach the limsup (i.e alpha) from above? We're saying that for large N it's less than or equal to alpha but i'm not sure why that is true
|a_n|^{1/n} may be greater than \alpha for finite numbers of n but not for infinite numbers of n. If for infinite number of values of n, |a_n|^{1/n} > \alpha, then lim sup would be greater than \alpha (because in that case there exists k such that the supremum of k-tail of the sequence would be greater than \alpha, contradiction).
Why can we remove the absolute values?
How do we know the supremum is greater than alpha?
It is not that we know the supremum is greater than alpha, is just beacause if x is a real number, observe that x
Weird proof of the last claim: If a series converges, but does not converge absolutely, alpha cannot be greater than 1 or less than 1 (by the first two claims), so it must be equal to 1. Since such series exist, it must be possible for a series with alpha equal to 1 to converge. And since it doesn't converge absolutely, and taking absolute values term by term obviously doesn't affect alpha, it is possible for a series with alpha equal to 1 to diverge. Proof by two examples I didn't bother to identify?
Your example additionally shows that a series with alpha equal to 1 could converge absolutely, though.
That series converge to a rational number. One question, does the slowly convergent series 1/(k*log(k)^2) converge to a transcendental? Every term is transcendental, so does the infinite sum converge to a transcendental?
Not necessarily
@@drpeyam It converges to approximately 2.1097, but how can we prove or disrpove that this number is transcendental. I forgot to say that in the sum, k goes from 2 to infinity.
@@guill3978 proving that a number is transience is very difficult as don't even know about e+pi
oeis.org/A115563 is the constant which I'm refering to. Is it transcendental?
Aditya Dwivedi If e + pi is algebraic, that means that the algebraics isn’t closed under addition. But the naturals, the integers, the rationals, even the Reals as a whole, are all closed under this operation. How does this work?
N̲ICE PF
.unseiged