Video: Determining the Convergence or Divergence of a Series Involving the Quotient between a Factorial Function and an Exponential Function

Consider the series βˆ‘_(𝑛 = 1) ^(∞) ((βˆ’1)^(𝑛) (2𝑛)!)/(3^(3𝑛)). Determine whether the series converges or diverges.

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Video Transcript

Consider the series the sum from 𝑛 equals one to ∞ of negative one to the 𝑛th power times two 𝑛 factorial divided by three to the power of three 𝑛. Determine whether the series converges or diverges.

We’re given a series and we need to determine the convergence or divergence of this series. And at this point, we’ve seen a lot of different methods to determine convergence or divergences of series. So the first thing we’re going to need to do is determine which of these methods would be the best for this series.

First things first, this series is an infinite series. It will have an infinite number of terms. One of the easier methods we could try is the 𝑛th-term divergence test. However, we can’t use this in this case because we don’t know what happens to two 𝑛 factorial divided by three to the power of three 𝑛 as 𝑛 approaches ∞. We get a similar story for the comparison test. It’s hard to think of a series which we can compare this to and which we know the convergence or divergence of. However, if we take a look at our summand, we can see it’s the quotient of two functions.

And normally, if our summand is the quotient of two functions, applying the ratio test is a good idea because often when we apply the ratio test, there’ll be a lot of cancelation which can make it a lot easier to deal with. So let’s start by recalling the ratio test for a series. For the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛, where we set 𝐿 equal to the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one divided by π‘Ž 𝑛, then the ratio test tells us if this value of 𝐿 is less than one, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 will converge absolutely.

However, the ratio test also tells us if this value of 𝐿 is greater than one, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 will be divergent. However, if this value of 𝐿 is equal to one, then the ratio test will be inconclusive. We can’t determine the convergence or divergence of this series from the ratio test. So we’ll set π‘Ž 𝑛 to be the summand of the series given to us in the question and now try applying the ratio test.

To apply the ratio test, we need to evaluate the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one divided by π‘Ž 𝑛. Before we start evaluating this limit, we’ll start by simplifying. Instead of evaluating π‘Ž 𝑛 plus one divided by π‘Ž 𝑛, we’ll instead multiply it by the reciprocal of π‘Ž 𝑛. This gives us the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one times the reciprocal of π‘Ž 𝑛. We’ll substitute 𝑛 plus one into our summand to find an expression for π‘Ž 𝑛 plus one. This gives us negative one to the power of 𝑛 plus one multiplied by two times 𝑛 plus one factorial all divided by three raised to the power of three times 𝑛 plus one.

And we can simplify this expression slightly. First, two multiplied by 𝑛 plus one is equal to two 𝑛 plus two. Next, in our denominator in the exponent, three times 𝑛 plus one can be simplified to give us three 𝑛 plus three. We then need to multiply this by the reciprocal of π‘Ž 𝑛. This gives us the limit as 𝑛 approaches ∞ of the absolute value of negative one to the power of 𝑛 plus one times two 𝑛 plus two factorial divided by three to the power of three 𝑛 plus three multiplied by three to the power of three 𝑛 divided by negative one to the 𝑛th power times two 𝑛 factorial.

We’re now ready to start simplifying. First, we have 𝑛 shared factors of negative one in our numerator and our denominator. We can cancel these out. This just leaves us with a factor of negative one in our numerator. Next, we have three 𝑛 shared factors of three in both our numerator and our denominator. So we can cancel these out. This just leaves us with three cubed in our denominator. Finally, by using the definition of a factorial, we know that two 𝑛 plus two factorial is equal to two 𝑛 plus two times two 𝑛 plus one multiplied by two 𝑛 factorial.

So we can cancel the shared factor of two 𝑛 factorial in our numerator and our denominator. So after all of these simplifications, we now have the limit as 𝑛 approaches ∞ of the absolute value of negative one times two 𝑛 plus two times two 𝑛 plus one all divided by three cubed. And at this point, we could keep simplifying to evaluate this limit. However, we can actually just evaluate this limit directly. Our limit is as 𝑛 is approaching ∞. Our denominator of three cubed remains constant; it doesn’t change as the value of 𝑛 changes.

The same is true for the factor in our numerator of negative one. However, we can see that both two 𝑛 plus two and two 𝑛 plus one are approaching ∞ as 𝑛 approaches ∞. So in this limit, our numerator is growing without bound; however, our denominator remains constant. This means this limit evaluates to give us positive ∞. And we might be worried at this point since the ratio test only tells us about the case where 𝐿 is less than one or where 𝐿 is greater than one. But in actual fact, the ratio test also holds true if the value of 𝐿 is positive ∞.

This means by using the ratio test, we can conclude that our series must be divergent. Therefore, by showing that the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms of the series given to us in the question approaches ∞, we were able to show by using the ratio test that this series must be divergent.

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