# Video: Using the Ratio Test to Determine Convergence

Consider the series β_(π = 0)^(β) π_(π), where π_(π) = π^(π)/((π + π)!) for some integers π, π > 1. (i) Calculate lim_(π β β) |π_(π + 1)/π_(π)|. (ii) Hence, decide whether the series converges or diverges.

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### Video Transcript

Consider the series, which is the sum from π equals zero to β of π π, where π π is equal to π to the π over π plus π factorial for some integers π and π, which are both greater than one. Part i, calculate the limit as π tends to β of the absolute value of π π plus one over π π. And part ii, hence decide whether the series converges or diverges.

For part 1, we simply need to find the limit as π tends to β if the absolute value of π π plus one over π π. Weβve been given π π in the question and itβs equal to π to the π over π plus π factorial. Therefore, we can say that π π plus one is equal to π to the π plus one over π plus one plus π factorial. So we can substitute in π π and π π plus one into our limit. So we have that our limit is equal to the limit as π tends to β of π π plus one. So thatβs π to the π plus one over π plus one plus π factorial multiplied by one over π π. So thatβs π plus π factorial over π to the π.

Immediately, we can see that we can do some cross cancelling here. We have π to the π plus one in the numerator of the fraction on the left. And we can cross cancel this with the π to the π in the denominator of the fraction on the right. In doing this, weβll simply be left with π in the numerator of the fraction on the left. Next, we can rewrite π plus one plus π factorial. π plus one plus π factorial is equal to π plus π plus one factorial. And this is also equal to π plus π plus one multiplied by π plus π factorial. And we can substitute this in to our fraction in our limit.

Weβre left with the limit as π tends to β of the absolute value of π multiplied by π plus π factorial over π plus π plus one multiplied by π plus π factorial. Therefore, this factor of π plus π factorial in the numerator and denominator can be cancelled. Weβre left with the limit as π tends to β of the absolute value of π over π plus π plus one. Here, weβre taking the limit as π tends to β. All the other terms in our fraction are constant. Since π appears in the denominator of the fraction and nowhere else, this means that the limit as π tends to β of this fraction will be equal to zero. This is because as π gets larger and larger and larger and closer to β, the fraction with π in the denominator will get closer and closer to zero. And so, we can say that this limit is equal to zero. Since weβve found the value of our limit, this concludes the first part of the question. Letβs now move on to part 2.

In part 2, weβ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 π tends to β of the absolute value of π π plus one over π π, this pushes us towards using the ratio test. Now, the ratio test tells us that for some series, which is the sum from π equals one to β of π π with πΏ being equal to the limit as π tends to β of the absolute value of π π plus one over π π, that if πΏ is less than one, then our series converges absolutely. If πΏ is greater than one, then our series diverges. And if πΏ 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 π equals one to β. However, our series goes from π equals zero to β. However, this does not matter. Weβre still able to apply the ratio test. This is because the sum from π equals zero to β of π π is equal to π nought plus the sum from π equals one to β of π π. So we can apply the ratio test to the sum from π equals one to β of π π. And in our case, π nought is equal to one over π factorial, where π is an integer which is greater than one. So we can see that π nought will be a constant. And so, the result of the ratio test on our series from π is equal to one will also apply to our series from π is equal to zero.

Letβs now apply the ratio test. In part 1, we found that the limit as π tends to β of the absolute value of π π plus one over π π is equal to zero. And so, we have that πΏ 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 π is equal to one to β of π π converges absolutely. We also have that π nought is finite. Hence, we can come to the conclusion that the sum from π is equal to zero to β of π π converges absolutely.