00:01
Okay, for this problem, we'll use the ratio test to figure out when we get convergence here.
00:08
For the ratio test, we limit as n goes to infinity of a .n, sorry, a .n plus 1 over a .n.
00:16
Where the a .n terms is this whole thing, so including the x values here.
00:22
So that's limit as n goes to infinity of absolute value of x minus 2 to the n plus 1, divided by n plus 1, squared plus 1.
00:38
So that's our a n plus 1 and we're dividing that by a n so multiplying by the reciprocal.
00:43
So we're multiplying by n squared plus 1 over x minus 2 to the n.
00:53
Okay so x minus 2 to the n divided by x minus 2 to the n is just going to be x minus 2 and then we have n squared plus 1 divided by n plus 1 squared plus 1.
01:18
So these up top and down below we have degree 2 polynomials with leading coefficient 1.
01:24
So as n goes to infinity, this is just going to go to 1.
01:28
So this is just going to turn out to be absolute value of x minus 2.
01:33
And we want for that to be less than 1.
01:38
Okay, so absolute value of x minus 2 is going to be less than 1 when x minus 2 is trapped between minus 1 and 1.
01:50
So if we have that minus 1 is less than x minus 2, we can add 2 to both sides and see that that corresponds to 1 is less than x.
02:02
And if we have x minus 2 is less than 1, we can add 2 to both sides to see that that corresponds to x is less than 3...