21. Find the interval of convergence of the power series....

21. Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval. If the answer is an interval, enter your answer using interval notation. If the answer is a finite set of values, enter your answers as a comma separated list of values.) sum_{n=0}^{infinity} n!(x - 9)^n 22. Find a geometric power series for the function, centered at 0. h(x) = 3 / (2 + x) sum_{n=0}^{infinity} ()

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00:02 Hi, here we are giving a power series n from 0 to infinity minus 1, let me zoom minus 1 to the n plus 1 times n plus 5 times x to the n and we're going to find the interval of convergence and we are also asked to check the for convergence at the end points of the interval.

00:28 Okay, so if we have a a power series is a series that it is of this form.

00:35 The infinite sum, a .n, x to the n, x is a variable.

00:40 And to find the radius of convergence, if we know the radius of convergence, the interval of convergence is minus rho, where rho is the radius of convergence.

00:51 To find this radius of convergence, what we do, we take the ratio of n, a .n plus 1 over a .n.

00:59 Then the absolute value of this then we take the limit as n goes to infinity.

01:04 Once we find the limit then we have that the radius of convergence is 1 over l.

01:10 This is in general what we have.

01:13 Of course it might be the case that the limit does not exist but this is not the case here.

01:18 So this is what we have over here and now let's see.

01:23 We have here in our series, a .n is given by minus 1 to the n plus 1.

01:31 Times n plus 5 so a n plus 1 over a n in the absolute value of this is going to be minus 1 to the n plus 2 times n plus 1 plus 5 over minus 1 to the n plus 1 over times excuse me n plus 5 so this is going to be now this because we take the absolute value it doesn't matter here just to say it briefly minus 1 to the n plus 2 over minus 1 to the n plus 1 this is equal to minus 1 but then we take the absolute value of this thing so this is equal to 1 and then of course we have that this term is positive this is positive so the whole thing is going to be equal to n plus 6 over n plus 5.

02:36 And this is equal to n times 1 plus 6 over n over n times 1 plus 5 over n.

02:48 This n and this end cancel out.

02:51 So this is equal to 1 plus 6 over 1 plus 5 over n.

02:58 And why am i doing this algebra? because i want to calculate the limit...

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