Determine whether the given series are convergent using the Ratio Test.∑n=1∞n!3n\sum_{n=1}^{\infty}\frac{n!}{3^{n}}

Diverges since L = ∞ > 1 L=\infty>1

Determine whether the given series are convergent using the Ratio Test. ∑ n = 1 ∞ n ! 3 n \sum_{n=1}^{\infty}\frac{n!}{3^{n}}

this problem, we're asked to determine whether the given series are convergent using the ratio test. Now I can see what we have is the series from n equals one to infinity of n factorial over three to the power of n. I can already tell since we have some exponentials and factorials here, the ratio test is indeed gonna be best. And since that's what we're asked to use, let's just go ahead and apply this. Now what I can see is that we have our a sub n right here. And keep in mind for the ratio test, you're also gonna want to find your a sub n plus one. So I'll just take these n's here and replace them with n plus one. So that gives us n plus one factorial over three to the n plus one. Now for the ratio test, we want to find l, which is the limit as n approaches infinity of the absolute value of a sub n plus one over a sub n. And so start this off, well, I'm going to be calculating this l here. So I have the limit as n approaches infinity, and we'll start with a sub n plus one. Well, a sub n plus one, that's going to be n plus one factorial over three to the n plus one. That's all going to be divided by a sub n, which is n factorial over three to the power of n. Now what I'm going to do from here is I'm going to try to simplify things inside the absolute value bars. And I can do that by flipping the bottom fraction and bringing it up to the top. So that's going to give me the absolute value of n plus one factorial over three to the n plus one, multiplied by three to the n, over n factorial. Now from here, what I need to do is try and simplify things even farther. And at this point, I want to see if I can get certain pieces to cancel here. Well, I can start by simplifying this part right here, n plus one factorial. And I can rewrite that as n plus one times n factorial since we're just going another step down and continuing this factorial. Now on bottom, three to the n plus one can also be written as three to the n times three to the one, which I can just write that three to the one as three. So all of this is going to be multiplied by three to the n over n factorial. Notice in this case the n factorials will cancel, and three to the n will cancel as well. So So what that's going to give me is the limit as n approaches infinity for the absolute value, and then on top we're just going to have n plus one. And then in the denominator, we're going to have three. So this is what we get. Now what's going to happen as n approaches infinity? Well, notice here the only n we have shows up in the numerator. And because this shows up in the numerator here, that means that this entire fraction is just going to blow up to infinity. That's what happens when we only have n in the numerator. So because of this, I can say that l is simply going to blow up to infinity. Now recall our rules when it comes to using the ratio test. When l is less than one, our series converges. When l is greater than one, our series diverges. And then when l is equal to one, our series is inconclusive. Now looking at this, l here is equal to infinity. And for the case where it diverges, we say that l is greater than one, but this also works for when l goes to infinity. So when l goes to infinity or l is greater than one, we'd say that our series diverges. So the solution to this problem is that the series is going to diverge. So So that's how you can solve problems like this using the ratio test. Hope you found this helpful, and let's move on.

Determine whether the given series are convergent using the Ratio Test. ∑ n = 1 ∞ n ! 3 n \sum_{n=1}^{\infty}\frac{n!}{3^{n}}

Diverges since L = ∞ > 1 L=\infty>1

Converges since L = − ∞ < 1 L=-\infty<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 are convergent using the Ratio Test.
$\sum_{n = 1}^{\infty} \frac{n!}{3^{n}}$

Converges since $L is less than 1$

Diverges since $L equals infinity is greater than 1$

Inconclusive since $L equals 1$

Converges since $L = - \infty < 1$

Verified step by step guidance

Step 1: Recall the Ratio Test. The Ratio Test states that for a series $ \sum_{n=1}^{\infty} a_n $, we calculate the limit $ L = \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} $. If $ L < 1 $, the series converges absolutely. If $ L > 1 $ or $ L = \infty $, the series diverges. If $ L = 1 $, the test is inconclusive.

Step 2: Identify the general term $ a_n $ of the series. Here, $ a_n = \frac{n!}{3^n} $.

Step 3: Compute the ratio $ \frac{|a_{n+1}|}{|a_n|} $. Substitute $ a_{n+1} = \frac{(n+1)!}{3^{n+1}} $ and $ a_n = \frac{n!}{3^n} $ into the ratio: \[ \frac{|a_{n+1}|}{|a_n|} = \frac{\frac{(n+1)!}{3^{n+1}}}{\frac{n!}{3^n}}. \]

Step 4: Simplify the ratio. Using the factorial property $ (n+1)! = (n+1) \cdot n! $, the ratio becomes: \[ \frac{|a_{n+1}|}{|a_n|} = \frac{(n+1) \cdot n!}{3^{n+1}} \cdot \frac{3^n}{n!} = \frac{n+1}{3}. \]

Step 5: Take the limit as $ n \to \infty $. The ratio $ \frac{n+1}{3} $ grows without bound as $ n \to \infty $, so $ L = \infty $. Since $ L > 1 $, the series diverges by the Ratio Test.

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