Determine whether the given series is convergent using the Ratio Test. ∑ n = 1 ∞ n 5 ( 6 ) n \sum_{n=1}^{\infty}\frac{n^5}{\left(6\right)^{n}}

this problem, we're asked to determine whether the given series are convergent using the ratio test. This is problem b that we're dealing with here, and in this problem, we have the sum from n equals one to infinity of five of n to the power of five over six to the n power. So how do we go about solving this? Well, we're specifically asked to use the ratio test when it comes to solving this problem. So let's see how we can do that. Well, I know for the ratio test, I first need to figure out what my a sub n is. I can see that's gonna be this piece that we have right here in front of the sum. And what I next need to do is find my a sub n plus one. A sub n plus one is gonna be our next step up, so I just replace these n's with n plus one. Meaning, we would have n plus one to the fifth power over six to the n plus one. Now why exactly am I doing this? Well, this is all because this is how we're going to use our ratio test. For the ratio test, we need to find l. We need to find l, which is gonna be the limit as n approaches infinity, and that's going to be four a sub n plus one over a sub n. So because this is what we're dealing with here, I can now find l since I have my a sub n plus one and my a sub n figured out. So we'll have the limit as n approaches infinity, and then on top, I'm gonna have my a sub n plus one. My a sub n plus one, I can see is this value, so we're going to have n plus one to the power of five over six to the n plus one. And what I'm going to do is divide this by my a sub n. And my a sub n, I can see it right here. That's gonna be n to the fifth power over six to the n. Now we have these absolute value bars right here, and what I can do is I can simplify this by flipping this fraction and bringing it up to the top. So we'll keep this limit as n approaches infinity, and we'll have the absolute value of n plus one to the fifth power over six to the n plus one. And all of this is going to be multiplied by the reciprocal of this fraction, which would be six to the n over n to the fifth power. Now from here, what I can do is I can simplify this six to the n plus one right here, because six to the n plus one, that's going to be the same thing as six to the n multiplied by six to the first power, which is just six to the n times six. Notice that the six to the n will now cancel completely. So what we're going to be left with here is the limit as n approaches infinity, and then we'll have the absolute value of n plus one to the fifth power on top divided by n to the fifth power on bottom inside these absolute value bars. Now what can I do from here? Well, I need to think about how n is going to behave as it approaches infinity based on what we have. Now on top here, we have n plus one to the fifth power. So the highest power we're going to have on top is an n to the fifth power, and on bottom, we also have an n to the fifth power. Since five is the highest exponent on both the ends, we take a ratio of the coefficients in front of the ends, which in this case is just going to be one over one, and one over one is the same thing as one. So we'll get that l is equal to one. Now because we got our l is equal to one, recall the rules when it comes to the ratio test. When l is less than one, our series converges. When l is greater than one, our series diverges. And when l is equal to one, the test is inconclusive. Now I notice for this example, we got a case where l is equal to one exactly. So because of that, we can say our series here is inconclusive for the ratio test. So we want to use some other kind of test to solve this problem. So we're not gonna go over a different type of test in this video because we were just asked to use the ratio test to determine convergence. But since we got inconclusive, that would be our final result. So I hope you found this helpful, and let's move on.

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 whether the given series is convergent using the Ratio Test.
$\sum_{n = 1}^{\infty} \frac{n^{5}}{{(6)}^{n}}$

Converges since $L equals 1$

Diverges since $L equals 1$

Inconclusive since $L equals 1$

Converges since $L is less than 1$

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 = n⁵ / 6ⁿ.

Step 4: Simplify the ratio. Combine terms and simplify: |aₙ₊₁ / aₙ| = |((n+1)⁵ / n⁵) * (1/6)|.

Step 5: Take the limit as n approaches infinity. Evaluate lim (n→∞) |aₙ₊₁ / aₙ|. Observe the behavior of (n+1)⁵ / n⁵ as n grows large. If the limit L = 1, the Ratio Test is inconclusive.

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 whether the given series is convergent using the Ratio Test.∑n=1∞n5(6)n\(\sum\)_{n=1}^{\(\infty\)}\(\frac{n^5}{\left(6\right)^{n}\)}

Watch next

Convergence Tests practice set

Determine whether the given series is convergent using the Ratio Test.
$\sum_{n = 1}^{\infty} \frac{n^{5}}{{(6)}^{n}}$