Consider the power series [ sumn=1^infty frac(-1)^n(3...

Consider the power series [ sum_{n=1}^{infty} frac{(-1)^{n}(3 x-4)^{n}}{sqrt{n+3}} ] Find the center and radius of convergence ( R ). If it is infinite, type "infinity" or "inf". Center ( a= ) ( 4 / 3 ) Radius ( R= ) ( 1 / 3 ) What is the interval of convergence? Give your answer in interval notation.

Transcript

00:01 Okay, to solve this problem, you already know that the center is 4 3rd, so why don't you get a number line? put 4 3rd.

00:08 The radius from the center to the radius would be your endpoints.

00:12 So if you go back 1 3rd, you'll get 4 3rd minus 1 3rd, giving us 1.

00:17 Now if we go the other way, 4 3rd plus 1 3rd is 5 3rd.

00:21 So our endpoints is 5 3rd and 1.

00:24 So that would be our interval of convergence.

00:27 Now you have to keep in mind that we need to check these endpoints.

00:31 Can we include them in the convergence? or are they actually divergent? so if they're divergent at those endpoints, we don't include it in the interval.

00:38 So when we test x equals to 1, we're going to plug it back into the series.

00:42 Series n equals 1 to infinity, negative 1 to the n, 3x, so 3 times 1 is 3, minus 4 to the n, square root of 1 plus 3.

00:53 So if we solve that, we would get negative 1 to the n times negative 1 to the n divided by square root of 4.

00:59 So negative 1n times negative 1n, these become positive.

01:02 So it would be 1 to the n, 2.

01:04 And i'll put the summation back.

01:07 Now if you think about it, what is going on here? well, you plug in values.

01:16 So we plugged in like 1 half to the 1.

01:19 And you're going to be adding 1 squared over 2 plus 1 cubed over 2 and so on.

01:30 Well, we're going to keep adding and adding and adding.

01:33 All we're just going to keep adding is 1 half plus 1 half plus 1 half plus 1 half forever.

01:41 Because 1 to the first, 1 to the square, 1 cubed is always 1 on top.

01:46 So you're just going to keep adding this.

01:47 You go 1, and then you get 1 and a half, and then you get 2.

01:51 So it's going to get bigger and bigger and bigger.

01:54 And that indicates to us that actually it's going to be divergent at x equals to 1.

02:02 Now let's check at x equals 5 third.

02:04 So when you plug in 5 third, you get summation n equals 1 to infinity.

02:09 You get negative 1 to the n, 3 times 5 third.

02:14 So this is nice.

02:15 The 3's will cancel.

02:16 So we're leaving with 5 minus 4 to the n.

02:19 And then afterwards, we'll get 5 thirds plus 3.

02:23 So this leaves us with negative 1 to the n times positive 1 to the n over...

02:29 We can just leave it instead as 3 and 5 thirds.

02:32 We don't have to solve for that.

02:34 So negative 1 n times positive 1 to the n.

02:37 The only difference is we're going to get negative 1 to the n together...

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