00:01
To figure out the radius of convergence, we'll use the ratio test here.
00:05
So take the limit as n goes to infinity, absolute value of a .n plus 1 over a .n.
00:13
And by a .n, we mean this whole thing.
00:15
So including the x values here.
00:19
So this is limit as n goes to infinity of absolute value of x plus 2 to the n plus 1 over 2 to the n plus 1, natural log of n plus 1.
00:35
So this is our an plus 1 term, dividing by a .n, so multiplying by the reciprocal.
00:42
So we're multiplying by 2n, natural log of n, divided by x plus 2 to the n.
00:53
So x plus 2 to the n plus 1 divided by x plus 2 to the n, that's just going to leave us with x plus 2.
01:04
2 to the n divided by 2 to the n plus 1 is going to leave us with 1 1⁄2.
01:08
And then we have ln of n divided by ln of n plus 1.
01:17
And to figure out the limit as n goes to infinity of natural log of n divided by natural log of n plus 1, you do lopatiles rule here.
01:30
So both the top and the bottom blow up to infinity.
01:34
So you can apply lopatiles rule.
01:37
Lopatiles rule, you do the derivative at the top, divide by the derivative at the bottom.
01:42
So we'd get 1 over n divided by 1 over n plus 1.
01:52
So we'd get limit as n goes to infinity, n plus 1 over n, which is just 1.
02:02
So ln of n over ln of n plus 1 does go to 1.
02:05
So this is just absolute value of x plus 2 over 2.
02:13
And we want for that to be less than one.
02:16
So if we multiply both sides by two, we get absolute value of x plus 2 is less than 2.
02:24
So at this point, you might already be able to see that the radius of convergence is 2.
02:29
If you can't, the way you figure it out is if this happens, then x plus 2 is trapped between minus 2 and positive 2.
02:39
So if minus 2 is less than x plus 2, if we subtract 2 from both sides, that means that minus 4 is less than x.
02:49
And if we have x plus 2 is less than 2, if we subtract 2 from both sides, we get that x is less than 0...