# Video: Using the Ratio Test to Determine Convergence

Consider the series β_(π = 0) ^(β) ((β2)^π)/(π3^(π + 1)). Calculate lim_(π β β) |(π_π + 1)/π_π|. Hence, determine whether the series converges or diverges.

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### Video Transcript

Consider the series the sum from π equals zero to β of negative two to the πth power over π times three to the power of π plus one. Calculate the limit as π approaches β of the absolute value of π sub π plus one over π sub π. Hence, determine whether the series converges or diverges.

Letβs look back at our original series. We know the general form is the sum of π π. So, weβre going to let π π be equal to the bit inside the sum. Itβs negative two to the power of π over π times three to the power of π plus one. We should now be able to see that in order to calculate the limit as π approaches β of the absolute value of π sub π plus one over π sub π, we need to work out π sub π plus one. To achieve this, we replace π with π plus one. So, the numerator becomes negative two to the power of π plus one. And then, the denominator becomes π plus one times three to the power of π plus one plus one or three to the power of π plus two.

We need to calculate the limit as π approaches β of the absolute value of the quotient of these. Now, if we were to write these in fractional form, it might get a bit confusing. So instead, we write it as the limit as π approaches β of the absolute value of negative two to the power of π plus one over π plus one times three to the power of π plus two divided by negative two to the πth power over π times three to the power of π plus one. Then, we recall that to divide by a fraction, we simply multiply by the reciprocal of that fraction. And so, our limit becomes the absolute value of negative two to the power of π plus one over π plus one times three to the power of π plus two times π times three to the power of π plus one over negative two to the πth power.

We also know that to divide two numbers whose base is equal, we simply subtract their exponents. So, negative two to the power of π plus one over negative two to the power of π is negative two to the power of π plus one minus π, which is negative two to the power of one or simply negative two. Similarly, three to the power of π plus one divided by three to the power of π plus two is three to the power of π plus one minus π plus two, which is three to the power of negative one or one-third. So, weβre simply left with three on the denominator. We can now rewrite our limit in a number of ways. Iβve chosen to separate negative two-thirds and π over π plus one. So, we want to calculate the limit as π approaches β of the absolute value of negative two-thirds times π over π plus one.

One of the laws of limits says we can take out any constant factors. Now, here weβre going to take out the constant factor of negative two-thirds. But of course, since this is the absolute value, we take out the absolute value of negative two-thirds, which is simply two-thirds. So, letβs find the limit as π approaches β of the absolute value of π over π plus one. And if we were to apply direct substitution right now, weβd get β over β, which is of indeterminate form. So instead, we perform some manipulation. We divide both the numerator and the denominator of our fraction by the highest power of π on our denominator. So, weβre going to divide both the numerator and the denominator by π. Letβs clear some space.

The fraction becomes π over π over π over π plus one over π. But of course, π over π is equal to one. And we need to calculate two-thirds of the limit as π approaches β of the absolute value of one over one plus one over π. Now, as π grows larger, one over π grows smaller. In other words, as π approaches β, one over π approaches zero. And so, the limit as π approaches β of the absolute value of one over one plus one over π becomes the absolute value of one over one, which is simply one. Two-thirds times one is two-thirds. And so, weβve calculated our limit. The limit as π approaches β of the absolute value of π sub π plus one over π sub π is two-thirds.

So, how do we use this to establish whether the series converges or diverges? Well, letβs say the limit as π approaches β of the absolute value of π sub π plus one over π sub π is equal to πΏ, the constant πΏ. The ratio test tells us that if πΏ is less than one, the series is absolutely convergent and hence converges. If itβs greater than one, the series diverges. And if itβs equal to one, our test is inconclusive. Well, our value for πΏ was two-thirds; this is quite clearly less than one. And so, by the ratio test, our series converges.