00:01
Okay so to find the interval convergence and the two endpoints of that interval, what we're going to do is use something called the root test.
00:12
So we can use limit n approaches infinity absolute value of the series.
00:18
So the series, i'm going to plug it in there, of the x minus 7 to the n, then to negative 6n to the n part.
00:29
You can take the absolute value once to the n.
00:31
So root test is limit n approaches infinity of absolute value a sub n to the 1 over n, where a of n is this.
00:44
So a power raised to a power, you can multiply all these.
00:48
So you can get limit n approaches infinity x minus 7, n over n times 1 divided by n is just 1, so just left with x minus 7.
00:58
And then this one's n to the 1 over n, and this one will be negative 6.
01:03
Right, once again there's power of 1, but you don't need to include it.
01:06
And then your absolute values, we're going to apply this limit to here because it's n approaches infinity.
01:15
So one of the common things to know is limit n approaches infinity, n raised to 1 over n, it's handy limit is just 1.
01:22
So you're going to be left with x minus 7 divided by negative 6 times 1, but negative 6 times 1 is just negative 6.
01:31
So we're going to leave it like that.
01:35
Now it is convergent, the interval of convergence, we know if this is true, that it'll be convergent when this is less than 1.
01:47
So what we're going to do is use that same inequality and set this to be less than 1.
01:53
Now you're left with absolute value x minus 7, the absolute value of negative 6 is 6, less than 1.
01:59
So you can multiply the 6 across and you'll have x minus 7 less than 6.
02:04
You can expand this out to be x minus 7 less than 6, and the other side will be negative 6.
02:08
Add 7 to both sides, and we figure out it's 1 between x, x is between 1 and 13.
02:16
So 1 and 13 would be your endpoints.
02:19
Now you need to check your endpoints to see if they're convergent at that exact point.
02:24
So what do i mean? well we can erase this first, and we're going to plug them in and see if the series is still convergent at those points.
02:31
So when x is 1, we're going to plug in 1 into the x.
02:35
You'll get summation n equals 1 to infinity, 1 minus 7 is negative 6 to the n, and then everything else is left alone.
02:45
Notice that these two will cancel, leaving you with summation n equals 1 to infinity, 1 over n.
02:52
This is actually a very common series, it's called a divergent harmonic series.
02:57
So we know it's divergent there.
03:01
So at x equals 1, it's divergent...