Ratio test, why does the (-1)^(n+1) disappear?

AI Thread Summary
In the ratio test, the term (-1)^(n+1) oscillates between -1 and 1, but it does not affect the limit as n approaches infinity. The test focuses on the absolute value of the ratio |An+1 / An|, which simplifies the calculations by allowing the oscillating terms to be disregarded. This is because the limit of the absolute values converges, while the oscillating signs do not impact the overall convergence behavior. Therefore, the oscillating factors can be factored out without affecting the outcome of the limit. The discussion clarifies that recognizing this allows for a simpler application of the ratio test.
CinderBlockFist
Messages
86
Reaction score
0
In the book while doing the ratio test:

as n --> infinity

why do they just have (-1)^(n+1) just disappear in the next step, since it oscillates between -1 and 1, I don't understand how u could just make it disappear in computations. Isn't the limit as n -> infinity , equal to D.N.E.?
 
Physics news on Phys.org
I believe the ratio test tests the absolute value of |An+1 / An|. so the (-1)^n and (-1)^n+1 will just be one, so u can just take it out.
 
crap lol, i didn't even think of that. THanks cyrus, u da man!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top