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?
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.
Yes, if it is true that |a_n+1 / a_n| > 1 eventually, then |a_n+1| > |a_n| eventually. This means that for large enough values of n, the terms of the sequence a_n are increasing in magnitude and so they cannot be approaching zero, which implies that the series of a_n diverges by the Divergence Test. Your intuition is spot on!
+Raul G Since |a_N+1| > L |a_N|, it naturally follows that L|a_N+1| > L^2 |a_N|. Applying this idea repeatedly allows us to show that the sequence a_n does not converge to 0 as the terms keep on getting larger, hence the series diverges.
thank you very much for proving a lot of infinite series' tests to the world
Thanks so much for this! It was very helpful and you're a great teacher.
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?
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.
Yes, if it is true that |a_n+1 / a_n| > 1 eventually, then |a_n+1| > |a_n| eventually. This means that for large enough values of n, the terms of the sequence a_n are increasing in magnitude and so they cannot be approaching zero, which implies that the series of a_n diverges by the Divergence Test. Your intuition is spot on!
@@slcmathpc thank you
@@slcmathpc Great, thanks both
Wonderful explanation
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.?
why did you multiply by L to get L^2 |a(n)|?
+Raul G Since |a_N+1| > L |a_N|, it naturally follows that L|a_N+1| > L^2 |a_N|. Applying this idea repeatedly allows us to show that the sequence a_n does not converge to 0 as the terms keep on getting larger, hence the series diverges.
hiii sir..
ratio test and alemberts ratio test both are same..? or different
Yes, they seem to be the same.
Thank you sir