Video: Using the Ratio Test to Determine Convergence

Consider the series βˆ‘_(𝑛 = 0)^(∞) π‘Ž_(𝑛), where π‘Ž_(𝑛) = 𝑏^(𝑛)/((𝑛 + 𝑐)!) for some integers 𝑏, 𝑐 > 1. (i) Calculate lim_(𝑛 β†’ ∞) |π‘Ž_(𝑛 + 1)/π‘Ž_(𝑛)|. (ii) Hence, decide whether the series converges or diverges.

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Video Transcript

Consider the series, which is the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛, where π‘Ž 𝑛 is equal to 𝑏 to the 𝑛 over 𝑛 plus 𝑐 factorial for some integers 𝑏 and 𝑐, which are both greater than one. Part i, calculate the limit as 𝑛 tends to ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛. And part ii, hence decide whether the series converges or diverges.

For part 1, we simply need to find the limit as 𝑛 tends to ∞ if the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛. We’ve been given π‘Ž 𝑛 in the question and it’s equal to 𝑏 to the 𝑛 over 𝑛 plus 𝑐 factorial. Therefore, we can say that π‘Ž 𝑛 plus one is equal to 𝑏 to the 𝑛 plus one over 𝑛 plus one plus 𝑐 factorial. So we can substitute in π‘Ž 𝑛 and π‘Ž 𝑛 plus one into our limit. So we have that our limit is equal to the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 plus one. So that’s 𝑏 to the 𝑛 plus one over 𝑛 plus one plus 𝑐 factorial multiplied by one over π‘Ž 𝑛. So that’s 𝑛 plus 𝑐 factorial over 𝑏 to the 𝑛.

Immediately, we can see that we can do some cross cancelling here. We have 𝑏 to the 𝑛 plus one in the numerator of the fraction on the left. And we can cross cancel this with the 𝑏 to the 𝑛 in the denominator of the fraction on the right. In doing this, we’ll simply be left with 𝑏 in the numerator of the fraction on the left. Next, we can rewrite 𝑛 plus one plus 𝑐 factorial. 𝑛 plus one plus 𝑐 factorial is equal to 𝑛 plus 𝑐 plus one factorial. And this is also equal to 𝑛 plus 𝑐 plus one multiplied by 𝑛 plus 𝑐 factorial. And we can substitute this in to our fraction in our limit.

We’re left with the limit as 𝑛 tends to ∞ of the absolute value of 𝑏 multiplied by 𝑛 plus 𝑐 factorial over 𝑛 plus 𝑐 plus one multiplied by 𝑛 plus 𝑐 factorial. Therefore, this factor of 𝑛 plus 𝑐 factorial in the numerator and denominator can be cancelled. We’re left with the limit as 𝑛 tends to ∞ of the absolute value of 𝑏 over 𝑛 plus 𝑐 plus one. Here, we’re taking the limit as 𝑛 tends to ∞. All the other terms in our fraction are constant. Since 𝑛 appears in the denominator of the fraction and nowhere else, this means that the limit as 𝑛 tends to ∞ of this fraction will be equal to zero. This is because as 𝑛 gets larger and larger and larger and closer to ∞, the fraction with 𝑛 in the denominator will get closer and closer to zero. And so, we can say that this limit is equal to zero. Since we’ve found the value of our limit, this concludes the first part of the question. Let’s now move on to part 2.

In part 2, we’re required to use our answer from part 1 to decide whether the series converges or diverges. Since in part 1, we calculated the limit as 𝑛 tends to ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛, this pushes us towards using the ratio test. Now, the ratio test tells us that for some series, which is the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 with 𝐿 being equal to the limit as 𝑛 tends to ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛, that if 𝐿 is less than one, then our series converges absolutely. If 𝐿 is greater than one, then our series diverges. And if 𝐿 is equal to one, then the test is inconclusive.

Now, the first thing we may notice is that the ratio test applies for series, which go from 𝑛 equals one to ∞. However, our series goes from 𝑛 equals zero to ∞. However, this does not matter. We’re still able to apply the ratio test. This is because the sum from 𝑛 equals zero to ∞ of π‘Ž 𝑛 is equal to π‘Ž nought plus the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛. So we can apply the ratio test to the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛. And in our case, π‘Ž nought is equal to one over 𝑐 factorial, where 𝑐 is an integer which is greater than one. So we can see that π‘Ž nought will be a constant. And so, the result of the ratio test on our series from 𝑛 is equal to one will also apply to our series from 𝑛 is equal to zero.

Let’s now apply the ratio test. In part 1, we found that the limit as 𝑛 tends to ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is equal to zero. And so, we have that 𝐿 is equal to zero. Zero is less than one. And so, we have satisfied part number 1 of the ratio test. This tells us that the sum from 𝑛 is equal to one to ∞ of π‘Ž 𝑛 converges absolutely. We also have that π‘Ž nought is finite. Hence, we can come to the conclusion that the sum from 𝑛 is equal to zero to ∞ of π‘Ž 𝑛 converges absolutely.

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