Use the Limit Comparison Test to determine whether the series converges.∑n=1∞n(n−1)3n+1\sum_{n=1}^{\infty}\frac{n}{\left(n-1\right)3^{n+1}}

Converges since L = 1 > 0 L=1>0 and b n b_{n} converges.

Use the Limit Comparison Test to determine whether the series converges. ∑ n = 1 ∞ n ( n − 1 ) 3 n + 1 \sum_{n=1}^{\infty}\frac{n}{\left(n-1\right)3^{n+1}}

this problem, we are asked to use the limit comparison test to determine whether each series converges. Now we already did example a in the previous video, so now we're going to do example b, where we have the sum from n equals one to infinity of n over n minus one times three to the n plus one. Now we're specifically asked to do a limit comparison test here. And whenever you're doing any kind of comparison test, a good place to start is by figuring out what the series in question is. Now I can see that's gonna be this a sub n right here. So now that I know my a sub n, I need to create another series, b sub n, that I can compare to my a sub n. Now how can I find a simpler series that I could compare? Well, I need to take a look at my a sub n and just think of ways I could make this more simple. If I'm trying to choose a b sub n, one thing I could do is take this negative one and knock it off here. Maybe that's gonna give me a more simple series. Well, let's find out. So doing that, I'm gonna get n on top over, and that's gonna be n, and then that negative one's gonna go away, so it's just gonna be n times three to the n plus one. Now from here, I can simplify, because I can cancel the n's right about there, and that's going to give me the series from n equals one to infinity of the sum, and that's going to be one over three to the n plus one. Now what I can do from here is I can recognize three to the n plus one is the the same thing as three to the n times three to the one or just three. And I can take this one third, and I can bring it out to the front. So this is going to be the same thing as the sum from n equals one to infinity of one third, which I'll pull out front of this this, this fraction here. And you could actually pull it to the front of the sum if you wanted to, but I'm gonna leave it inside the sum. And I'll show you why in a moment right here. So that's gonna be one third times one over three to the n power. Now right at this point, it might look like there's not something straightforward that we could do to to determine whether or not this series converges. But recognize that one is the same thing as one to the n, and then we have a three to the n on bottom. That's because one raised to a power of n would still just give us one. So I could rewrite this whole thing as one third to the n. Now notice writing it like this, we have a geometric series. And in this case, this right here would be the r. So what am I going to use for this? Well, what I can do is remember that we need to check the absolute value of r. Which is gonna be the absolute value of one third. Now notice here that one third is less than one. Now we learned for a geometric series, when this happens, the series is going to converge. So we can conclude that our series b sub n here converges, where our b sub n is what we chose right about here. We also have it over here. So this was our b sub n. Now what I need to do from here, now that I figured out that b sub n converges, is I want to go through this process of doing the limit comparison test. And the way that I can do that is by finding the limit as n approaches infinity for the ratio of a sub n over b sub n. So that's gonna be n over n minus one times three to the n plus one, and then this whole thing is going to be over my b sub n. Well, my b sub n, I can see is right here and right there, but I'm gonna go ahead and use this version. I think that's gonna be a bit more simple. So I have n over n times three to the n plus one. Now what I can do is I can go ahead and flip this bottom fraction and bring it up to the top. So I have the limit as n approaches infinity, then it's going to be n over n minus one times three to the n plus one multiplied by, and then flipping this, we'll have n times three to the n plus one over n. Now writing it like this, notice that the n's are going to cancel right here. So we can go ahead and cancel the n's here, and then I can also cancel three to the n plus one. So all I'm going to be left with for this limit as n approaches infinity is n over n minus one. And I noticed that my highest exponent for n on on top is just n to the first power, and it's the same n to the first power on bottom. So So because our highest exponents are both n to the first power, I need to create a ratio between my n's here, the coefficients out in front. So that's gonna give me one over one, which is just one. So we can conclude that our l is equal to one. Now remember, when using the limit comparison test, whenever l is greater than zero, well, in this case, what we want to do is look at b sub n because we know that this is going to match how our b sub n behaved. And I can see here that our b sub n converges. So because our b sub n converges, this was our b sub n right here, then I conclude I can conclude that our series a sub n is also going to converge. So this right here would be the solution to example e. Hope you found this helpful, and let's move on and get some more practice.

Use the Limit Comparison Test to determine whether the series converges. ∑ n = 1 ∞ n ( n − 1 ) 3 n + 1 \sum_{n=1}^{\infty}\frac{n}{\left(n-1\right)3^{n+1}}

Converges since L = 1 > 0 L=1>0 and b n b_{n} converges.

Diverges since L = − 1 < 0 L=-1<0 and b n b_{n} b n diverges.

Diverges since L = 1 > 0 L=1>0 L = 1 > 0 and b n b_{n} b n converges.

Converges since L = − 1 < 0 L=-1<0 L = − 1 < 0 and b n b_{n} b n converges.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Use the Limit Comparison Test to determine whether the series converges.
$\sum_{n = 1}^{\infty} \frac{n}{(n - 1) 3^{n + 1}}$

Converges since $L equals 1 is greater than 0$ and $b_{n}$ converges.

Diverges since $L = - 1 < 0$ and $b_{n}$ diverges.

Diverges since $L=1>0$ and $b_{n}$ converges.

Converges since $L=-1<0$ and $b_{n}$ converges.

Verified step by step guidance

Step 1: Identify the given series and rewrite it in a simplified form. The series is $ \sum_{n=1}^{\infty} \frac{n}{(n-1)3^{n+1}} $. To apply the Limit Comparison Test, we need to compare this series to a simpler series $ b_n $ that has a known convergence behavior.

Step 2: Choose a comparison series $ b_n $. A good choice is $ b_n = \frac{1}{3^n} $, because the exponential term $ 3^n $ dominates the growth of the denominator as $ n \to \infty $. This series $ \sum_{n=1}^{\infty} \frac{1}{3^n} $ is a geometric series with a common ratio $ r = \frac{1}{3} $, which is known to converge.

Step 3: Compute the limit $ L = \lim_{n \to \infty} \frac{a_n}{b_n} $, where $ a_n = \frac{n}{(n-1)3^{n+1}} $ and $ b_n = \frac{1}{3^n} $. Substitute $ a_n $ and $ b_n $ into the formula for $ L $: $ L = \lim_{n \to \infty} \frac{\frac{n}{(n-1)3^{n+1}}}{\frac{1}{3^n}} $. Simplify the expression.

Step 4: Simplify the limit expression. Combine terms and cancel out $ 3^n $ where possible: $ L = \lim_{n \to \infty} \frac{n \cdot 3^n}{(n-1) \cdot 3^{n+1}} $. This simplifies to $ L = \lim_{n \to \infty} \frac{n}{(n-1) \cdot 3} $. Factor and analyze the behavior of $ n $ and $ n-1 $ as $ n \to \infty $.

Step 5: Determine the value of $ L $. If $ L > 0 $ and finite, the Limit Comparison Test states that the convergence behavior of $ \sum a_n $ matches that of $ \sum b_n $. Since $ \sum b_n $ converges, $ \sum a_n $ also converges. Conclude the result based on the computed $ L $.

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