For (2), use the Cauchy Condensation Test, which states: given the sequence a(n), Σa(n) conv. iff Σ2ⁿa(2ⁿ) conv. So for (2):Σ ln²(n)/n² conv. iff Σ2ⁿ[ln²(2ⁿ)/2²ⁿ] ≡ Σn²ln²2/2ⁿ which conv. by Root Test; thus (2) converges. Alternatively, it can be shown that since n^¼ > ln(n) for n>>1, n^½> ln²(n) for n>> 1. From this, use the comparison test to show that since (2) < Σ n^½/n² = Σn^-1.5 which conv. by the p-series test, therefore (2) converges.
I have another way to solve the second infinite serie. n^(3/2) * ln(n)^2/n^2 = ln(n)^2/n^(1/2) which tends to 0. So it will be less than A (a number). (ln(n)/n)^2 < A * 1/n^(3/2) and we know that the serie on the right converges. So we have done it. This Riemann test is a good use for convergence or divergence.
Can you make more videos on limits and their definitions? I am starting a Calculus 1 course and we are studying the limit of a sequence. It would be nice to see videos on all different limit definitions and proofs
Since the riemann zeta function zeta(s) is twice differentiable at s = 2, the series sum ln(n)^2 / n^2 = zeta''(2) exists. Similarly since zeta(s) has a pole of order 1 at s = 1, sum ln(n)/n doesn't exist since loosely speaking it would be -zeta'(1).
5:38
Isn’t that supposed to be a negative?
Ah yes. Thanks for pointing that out.
For (2), use the Cauchy Condensation Test, which states: given the sequence a(n), Σa(n) conv. iff Σ2ⁿa(2ⁿ) conv.
So for (2):Σ ln²(n)/n² conv. iff Σ2ⁿ[ln²(2ⁿ)/2²ⁿ] ≡ Σn²ln²2/2ⁿ which conv. by Root Test; thus (2) converges.
Alternatively, it can be shown that since n^¼ > ln(n) for n>>1, n^½> ln²(n) for n>> 1. From this, use the comparison test to show that since (2) < Σ n^½/n² = Σn^-1.5 which conv. by the p-series test, therefore (2) converges.
I have another way to solve the second infinite serie.
n^(3/2) * ln(n)^2/n^2 = ln(n)^2/n^(1/2) which tends to 0.
So it will be less than A (a number).
(ln(n)/n)^2 < A * 1/n^(3/2) and we know that the serie on the right converges.
So we have done it.
This Riemann test is a good use for convergence or divergence.
For the square one, you can use the gamma/pi function to see it's less than 2! because the bottom limit is greater than 0, hence convergant.
Can you make more videos on limits and their definitions? I am starting a Calculus 1 course and we are studying the limit of a sequence. It would be nice to see videos on all different limit definitions and proofs
I'd like to be in a class of mr. BPRP...
Since the riemann zeta function zeta(s) is twice differentiable at s = 2, the series sum ln(n)^2 / n^2 = zeta''(2) exists. Similarly since zeta(s) has a pole of order 1 at s = 1, sum ln(n)/n doesn't exist since loosely speaking it would be -zeta'(1).
That's some serious overkill. That's like using the quadratic formula for x²-a=0 or Group theory to solve a permutations problem.
Happy Diwali my loving Sir....
💘 From India
I didn’t know how to solve the other 2 😂
Luckily the first one failed the integration test
can you prove those tests ?
Hello Kirby
I've been a practicing engineer for about 30 years now. I never see this stuff in the engineering world. Maybe the science guys see it. I don't.
dominated convergence FTW