Determine the convergence or divergence of the series. ∑ n = 1 ∞ n ∙ 8 n ( n + 1 ) ! \sum_{n=1}^{\infty}\frac{n\bullet8^{n}}{\left(n+1\right)!}

this problem, we are asked to determine the convergence or divergence of the following series. Now for this series, we have the sum from n equals one to infinity of n times eight to the n power over n plus one factorial. How do I go about solving this? Well, notice that we are not given what tests we need to use here, so we're going to have to figure that out. And I specifically chose to not provide the flowchart in this video. Now, here's what I recommend doing. I would suggest try solving this problem on your own without the flowchart and see if you can just recognize what type of series we're dealing with and how we can go about using a proper test to solve this. Now if you find yourself getting stuck at all, feel free to use the flowchart. But it's important that we can understand how to quickly work through these series when we see them. So let's just go ahead and jump right in. Now looking at this, a first thing that I might try, since I'm not really seeing anything that's super familiar here, is to do a divergence test. So a divergence test would be taking the limit as n approaches infinity for a sub n. Now a sub n is gonna be this whole piece right here. Now there's something that I notice about a sub n. I notice that we have this factorial in the denominator. We also have an exponential in the numerator, but the factorial grows faster than the exponential does when you have a limit as n approaches infinity. So what that means is this denominator is going to blow up faster, setting this whole fraction to zero. So what that means is that our limit here is going to come out to zero, which means our divergence test would be inconclusive. So now we need to think about a different test that we can use. And it might be a little bit hard initially to tell what we could do here. I can see that we don't have any kind of, like, negative one to a power or any sines or cosines, so it's not gonna be an alternating series, and this doesn't look like any of our special cases either. I don't really see this being in the form of some kind of geometric series, nor do I see a case where we have a p series or harmonic series since we have this factorial in the denominator. But there is something that comes to mind here. Because I see a factorial and I see an exponential, I recall there is a certain test we like to use when we see these showing up in our series. And the test that we use is the Ratio Test. So recall, for the ratio test, what we need to do is take a limit as n approaches infinity. And the limit is going to be of the absolute value of a sub n plus one over a sub n, and this whole thing is equal to r. R is not the ratio, it's the limit, but it tells us about our convergence here. So let's go ahead and go through the ratio test to see if this gives us an answer. So for the ratio test, what I need to do is figure out what my a sub n plus one is. A sub n plus one is gonna be taking all the n's that I see here and replacing them with n plus one. So we'll have n plus one times eight to the n plus one power over n plus one plus one, which always just come out to n plus two factorial. So that's my a sub n plus one. Now what I can do is use this to solve what we're dealing with down here. So we have the limit as n approaches infinity. This is gonna be of the absolute value of a sub n plus one. Now a sub n plus one is the first thing, so it's gonna be n plus one times eight to the n plus one over n plus two factorial. Now what we need to do is divide this whole thing by a sub n. But rather than dividing this fraction by another fraction, I'm just gonna flip this fraction and multiply it. That's the rule that we should know about. So in this case, what that means is I'm gonna flip my a sub n, and doing this is gonna give me n plus one factorial in the numerator and then n times eight to the n in the denominator. Now what I need to do from here is think of a way that I can simplify this whole thing, And I can think of a way to do that by manipulating this left piece here. So what I'm gonna do, I'm gonna continue on with this limit. You wanna make sure to include this. And then we're gonna have the absolute value, and what I'm gonna do is take this n plus one. We're gonna keep that. And eight to the n plus one is the same thing as eight to the n times eight to the one. And I'll leave the eight to the one as just eight. And then the denominator, I'm gonna take this n plus two factorial, I'm gonna replace it with n plus two times n plus one factorial. That's how factorials work, we just keep going down by one and multiplying it by the next by the previous term, and then what this is gonna be is multiplied by everything that we have on the right side here. So it's gonna be n plus one factorial over n times eight to the n. Now what do I do from here? We'll notice here that the n plus one factorial, now that we expanded this out on bottom here, will cancel, and the eight to the n also cancels. So what that's gonna leave me with on top is everything that we have left over. So I'll go ahead and I'll keep this limit right here, and we'll keep going. And so on top, we're going to have eight times n plus one, which if I go ahead and distribute that eight in there, well, that's gonna give me eight n plus eight. And then in the denominator, I can take this n and distribute it in here, giving me n squared plus two n. Now this right here is the limit that we're dealing with. And what I wanna do from here is look for my highest exponent on n. You can see the highest exponent we have in the numerator is n to the one power, and the highest exponent in the denominator is n squared. Notice here since we have a higher exponent on bottom for n, that means the bottom is going to be growing faster than the top. And as the bottom goes to infinity, this whole thing is just going to go to zero. So what that means is that this limit is going to come out to zero. So this right here is the r value. Now what can I say about this r right here? Well, recall how this works when it comes to the ratio test. If r is less than one, then we would say that our series converges. If r is greater than one, then we say that our series diverges. And if r is equal to one, then that means the test is inconclusive. But what I can clearly see here, based on the problem that we have, is that we've got an r value of zero. And zero, I can see, is clearly less than one. So because of that, I can conclude that our series converges based on the ratio test. So hope you found this video helpful, and let's move on and get some more practice.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!

Multiple Choice

Determine the convergence or divergence of the series.
$\sum_{n = 1}^{\infty} \frac{n ∙ 8^{n}}{(n + 1)!}$

Inconclusive

Diverges

Converges

Verified step by step guidance

Step 1: Recognize that the series is given as $ \sum_{n=1}^{\infty} \frac{n \cdot 8^n}{(n+1)!} $. To determine convergence or divergence, we can use the Ratio Test, which is effective for series involving factorials and exponential terms.

Step 2: Apply the Ratio Test. The Ratio Test states that for a series $ \sum a_n $, if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L $, then: (a) if $ L < 1 $, the series converges; (b) if $ L > 1 $, the series diverges; (c) if $ L = 1 $, the test is inconclusive.

Step 3: Compute $ \frac{a_{n+1}}{a_n} $ for the given series. Substitute $ a_n = \frac{n \cdot 8^n}{(n+1)!} $ and $ a_{n+1} = \frac{(n+1) \cdot 8^{n+1}}{(n+2)!} $. Simplify the ratio $ \frac{a_{n+1}}{a_n} $: $ \frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot 8^{n+1} \cdot (n+1)!}{(n+2)! \cdot n \cdot 8^n} $.

Step 4: Simplify the expression further. Factor out $ 8^{n+1} $ and $ 8^n $, and simplify the factorial terms $ (n+2)! $ and $ (n+1)! $. This leads to $ \frac{a_{n+1}}{a_n} = \frac{8 \cdot (n+1)}{n+2} $.

Step 5: Take the limit as $ n \to \infty $: $ \lim_{n \to \infty} \frac{8 \cdot (n+1)}{n+2} $. Simplify the expression to find $ L $. If $ L < 1 $, the series converges; if $ L > 1 $, it diverges. Use this result to conclude the behavior of the series.

5:44m

Watch next

Master Divergence Test (nth Term Test) with a bite sized video explanation from Patrick

Convergence Tests practice set

Problem sets built by lead tutors

Expert video explanations

Struggling with Calculus?

Determine the convergence or divergence of the series. ∑n=1∞n∙8n(n+1)!\sum_{n=1}^{\infty}\frac{n\bullet8^{n}}{\left(n+1\right)!}

Watch next

Convergence Tests practice set

Determine the convergence or divergence of the series.
$\sum_{n = 1}^{\infty} \frac{n ∙ 8^{n}}{(n + 1)!}$