Determine whether the given series is convergent using the Ratio Test.∑n=1∞8n(2n)!\sum_{n=1}^{\infty}\frac{8^{n}}{\left(2n\right)!}

Converges since L = 0 < 1 L=0<1

Determine whether the given series is convergent using the Ratio Test. ∑ n = 1 ∞ 8 n ( 2 n ) ! \sum_{n=1}^{\infty}\frac{8^{n}}{\left(2n\right)!}

this problem, we're asked to determine if the given series are convergent using the ratio test, and we'll start right here with problem a. Here we have the sum from n equals one to infinity, and this is gonna be eight to the power of n over two n factorial. Now just because I have these exponentials and factorials, this is already a good sign I would need to use the root the ratio test. But since we actually are told to use the ratio test, that's what I'm gonna use. So what I first want to do is figure out what my a sub n is, and that's gonna be all of this right here. Now for the ratio test, we need to take the limit as n approaches infinity for a sub n plus one over a sub n. And this is gonna be inside these absolute value bars. Now since I know what my a sub n is, I can figure out my a sub n plus one. This just means taking all of my n's and replacing them with n plus one. So this can give me eight to the n plus one power on top divided by, and then we'll have two times n plus one. And that's going to be factorial right there. Now I'm going to simplify this a little bit, because what I can do is distribute this two into the parentheses. That's going to give me two n plus two, and all of this is going to be within our factorial. So So this is what we get. So now we've gone ahead and found our a sub n and our a sub n plus one. And so now I can go ahead and apply this test right here, which is going to give me my l. So l is going to be equal to the limit as n approaches infinity. That's going to be the absolute value of a sub n plus one, which is eight to the n plus one power over two n plus two factorial, and all of this is going to be divided by a sub n. So it's going to be eight to the power of n over 2n factorial within these absolute value bars. And what do I do from here? Well, what I'm going to do is simplify what we have inside the absolute value by taking the reciprocal of this bottom fraction here and flipping it and bringing it to the top. It's going to be the absolute value of eight to the n plus one over two n plus two factorial. And then what we'll have here is that's multiplied by the reciprocal, which is going to be two n factorial over eight to the power of n. Now what I can do is try to simplify things even farther by changing what we have on the left side here. I'm not not gonna change the value of that fraction right there. I'm just going to see if I can modify what we have to get things to cancel on the right side here. So I'll still write an equivalent expression. So eight to the n plus one is the same thing as eight to the n times eight to the one, which is just eight to the n times eight. And then two n plus two factorial, well, that's gonna be the same thing as two n plus two times two n plus one times two n, and that's just gonna keep going because that's how the factorial works. That's gonna be multiplied by two n factorial over eight to the power of n, in which case I can see right here, the eight to the n is going to cancel, and the two the two n factorial will cancel as well. So what that's going to leave me with is the limit as n approaches infinity, and it's going to be four eight, the absolute value, over and it's gonna be two n plus two times two n plus one. Now what I could do is use the foil method to distribute everything out on bottom, but I actually don't need to worry about this since I can see the only n's that show up here are in the denominator, and we have the limit as n approaches infinity. We know when the denominator pulls up to infinity, the entire fraction is just going to go to zero. So because of that, our l is all going to come out to zero. So that's what we get when we find l. Now recall the rules when it comes to dealing with the ratio test. When l is less than one, the series converges. When l is greater than one, the series diverges. And then when l is equal to one, the series is inconclusive. Now looking at this, I can see that l is zero, which is clearly less than one. So because of that, we can conclude the series converges, and that would be the solution to the problem. So in this case, our series converges using the ratio test. Hope you found this helpful, and let's move on.

Determine whether the given series is convergent using the Ratio Test. ∑ n = 1 ∞ 8 n ( 2 n ) ! \sum_{n=1}^{\infty}\frac{8^{n}}{\left(2n\right)!}

Converges since L = 0 < 1 L=0<1

Diverges since L = 0 < 1 L=0<1 L = 0 < 1

Converges since L = 8 > 1 L=8>1

Diverges since L = 8 > 1 L=8>1 L = 8 > 1

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Determine whether the given series is convergent using the Ratio Test.
$\sum_{n = 1}^{\infty} \frac{8^{n}}{(2 n)!}$

Converges since $L equals 0 is less than 1$

Diverges since $L=0<1$

Converges since $L equals 8 is greater than 1$

Diverges since $L=8>1$

Verified step by step guidance

Step 1: Recall the Ratio Test. The Ratio Test states that for a series ∑aₙ, if the limit L = lim (n → ∞) |aₙ₊₁ / aₙ| exists, then: (a) If L < 1, the series converges absolutely. (b) If L > 1, the series diverges. (c) If L = 1, the test is inconclusive.

Step 2: Identify the general term aₙ of the series. Here, aₙ = 8ⁿ / (2n)!.

Step 3: Compute the ratio |aₙ₊₁ / aₙ|. Substitute aₙ₊₁ = 8ⁿ⁺¹ / (2(n+1))! and divide it by aₙ = 8ⁿ / (2n)!. Simplify the expression carefully.

Step 4: Simplify the factorial term (2(n+1))! in the denominator. Recall that (2(n+1))! = (2n+2)(2n+1)(2n)! and use this to simplify the ratio.

Step 5: Take the limit as n → ∞ of the simplified ratio. Analyze the behavior of the terms and determine whether the limit L is less than, greater than, or equal to 1. Use this result to conclude whether the series converges or diverges.

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