00:01
In this question, we are asked to find the radius of convergence and the interval of convergence of the given power series.
00:07
To do that, we are going to use the ratio test.
00:12
By the ratio test, we need to calculate the limit of the absolute value of a .n plus 1 over a .n.
00:20
Where a .n is the general term of the series.
00:27
So this is a .n.
00:29
All right.
00:30
The limit equals to the limit as n goes to infinity.
00:36
And to get a .n plus 1, we simply need to replace n by a.
00:39
N plus 1 everywhere in the formula for a n.
00:42
We are going to get 6 to the n plus 1 times x plus 6 to the n plus 1 divided by the square root of n plus 1.
00:53
That's a n plus 1 divided by a.
00:56
6 to the n x plus 6 to the n, x plus 6 to the n divide by the square root of n.
01:07
Now let's get rid of this ugly fraction.
01:12
So recall that to divide two fractions, we need to multiply the fraction in the numerator by the reciprocal of the fraction in the denominator.
01:49
We can cancel now 6 to the n and we can cancel x plus 6 to the n.
01:55
The limit would simplify to the absolute value of 6 times x plus 6 multiplied by the square root of n over n plus 1.
02:15
And when n goes to infinity, n over n plus 1 goes to 1.
02:23
So in the limit we're going to get 6 times the absolute value of x plus 6.
02:30
And if we want the series to converge, we want this limit to be less than 1.
02:38
Therefore, we want the absolute value of x plus 6 to be less than 1 or 6.
02:47
This means that x plus 6 is less than 1 over 6 and greater than negative 1 or 6.
03:00
Or x is less than 1 over 6 minus 6 and greater than negative 1 over 6 minus 6...