00:01
In this question, we are asked to find the interval of convergence of the given power series.
00:06
And to do that, we will use the ratio test.
00:11
By the ratio test, we need to calculate the limit of the absolute value of an plus 1 over an, where an plus an is the general term of the series.
00:24
So this is an.
00:24
And to get an plus 1, we simply need to replace n by n plus 1 in the formula for the general term of the series.
00:25
So this is an.
00:27
And to get an plus 1, we simply need to replace n by n plus 1 in the formula for the general.
00:36
So this is what we are going to get.
00:59
And we need to calculate the limit of the absolute value as n goes to infinity.
01:04
Let's simplify the expression.
01:08
We can rewrite that fraction as x minus 3 to the n plus 2 over n plus 2 times 4 to the n plus second power multiplied by the reciprocal of the denominator.
01:32
Then we can do some cancellations.
01:35
We can cancel 4 to the n plus 1 power.
01:38
And we can cancel x minus 3 to the n plus first and we are going to get the limit of the absolute value of x minus 3 multiplied by n plus 1 over n plus 2 and as n goes to infinity n plus 1 over n plus 2 goes to 1 and in the limit we are going to get the absolute value of x minus 3 and by the ratio test, the series converges if that limit is less less than one.
02:19
So the series converges for the absolute value of x minus 3 less than 1.
02:23
However, the ratio test doesn't tell us what happens when the limit equals to 1.
02:29
So we have to check that case separately.
02:33
We have to check what happens when x minus 3 equals to negative 1 and x minus 3 equals to 1.
02:43
We need to test the series directly in these two cases...