The series of 1/n^2 does converge by the p-series test, which I used implicitly with the Direct Comparison Test to show 19/8 is included in the interval of convergence. However, you can't use the p-series test directly on an alternating series (which is what you get when you plug in the other endpoint x = 21/8) since the p-series test is designed only for positive-term series. At any rate, I would argue even thinking about p-series is overkill for an alternating series: if a series is alternating (and the terms decrease in magnitude to zero), all you need to verify is the limit of the terms is zero.
At 1:03:04, why couldn't you just automatically say that when testing x=3, 1/ln(n) is convergent since we said that 1/ln(n) was convergent when testing x=-4?
Because in the case of x = -4, we had an alternating series, and the Alternating Series Test says it converges if the positive version of the terms converge to zero. But in the case of x = 3, the series is NOT alternating, so we can't use AST, so we have to use some other test, like the Direct Comparison Test, which tells us the series diverges in this case.
wow this guy sounds like my math professor
the one dislike is from joe khalig's fanbase smh
Can we say that the stuff around 51:50 is convergent by p-series test, since (1/(n^2)) => p=2 => p>1 => converges?
The series of 1/n^2 does converge by the p-series test, which I used implicitly with the Direct Comparison Test to show 19/8 is included in the interval of convergence. However, you can't use the p-series test directly on an alternating series (which is what you get when you plug in the other endpoint x = 21/8) since the p-series test is designed only for positive-term series. At any rate, I would argue even thinking about p-series is overkill for an alternating series: if a series is alternating (and the terms decrease in magnitude to zero), all you need to verify is the limit of the terms is zero.
At 1:03:04, why couldn't you just automatically say that when testing x=3, 1/ln(n) is convergent since we said that 1/ln(n) was convergent when testing x=-4?
Because in the case of x = -4, we had an alternating series, and the Alternating Series Test says it converges if the positive version of the terms converge to zero. But in the case of x = 3, the series is NOT alternating, so we can't use AST, so we have to use some other test, like the Direct Comparison Test, which tells us the series diverges in this case.
replay 32:25