The Ratio Test - Proof of Part (a)

[stevenkaban]

thank you very much for proving a lot of infinite series' tests to the world

[saisai]

Thanks so much for this! It was very helpful and you're a great teacher.

[riley]

thank you for this proof. I was wondering also as a possible shorter way of doing this, couldn't you just observe that since the ratio between the next term and the term before it is greater than 1, then each term is greater than the one before it. Therefore the sequence must tend to infinity because the terms are increasing. And if you have negative terms by an alternating series, then if the absolute value of the ratio is larger than one, then the sequence still cannot go to zero because each term is increasing in magnitude, so it must diverge. Does this work? Thanks.

[JnSubli]

Hi, My understanding for series to diverge is when "nth divergence test" must also meet this criteria lim n→∞ aₙ = lim n→∞+1 aₙ lim n→∞+∞ aₙ ≠ 0 else they would either Diverge (don't summed up to a real number) or Undefined (if all summed up to zero and infinitely repeating). Example for aₙ terms that are lim n→∞ (-1)^n or lim n→∞ sin(n) is Undefined. And for your case above lim n→∞ |aₙ| = ∞ ⇒ lim n→∞ aₙ ≠ 0 it may not meet lim n→∞ aₙ = lim n→∞+1 aₙ because lim n→∞ aₙ and n→∞+1 aₙ could also be +- aside from ++ or -- according to the ratio test |aₙ₊₁/aₙ| > 1. It Diverge only because the terms didn't summed up to zero. What do you think sir?

[derickblacido]

don´t understand why does the lim_n=>oo f(x) tend to "n", if the maximum value that the lim can take is just k.?

[ladedadedaschlobonmeknob]

why did you multiply by L to get L^2 |a(n)|?

[shivashivarathri]

hiii sir..
ratio test and alemberts ratio test both are same..? or different