Root Test Proof

In this video, I prove the root test, which is a classical and powerful test to determine if a series converges or not.

[NovayaK]

What a mad coincidence. I just got off my Calc BC class from learning about the Root Test.

[puerulus]

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.).

[prashantsharma-mchh]

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 <= lim sup so this proves lim sup >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).

[annali]

omg I just need it, like literally right now

[elfmas]

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

[Happy_Abe]

Why can we remove the absolute values?
How do we know the supremum is greater than alpha?

[iabervon]

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.

[prashantsharma-mchh]

Which book's section 11 are you referring ? @ 17:17

[guill]

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?