Video: Using the Ratio Test to Determine Convergence

Consider the series βˆ‘_(𝑛 = 0) ^(∞) ((βˆ’2)^𝑛)/(𝑛3^(𝑛 + 1)). Calculate lim_(𝑛 β†’ ∞) |(π‘Ž_𝑛 + 1)/π‘Ž_𝑛|. Hence, determine whether the series converges or diverges.

03:59

Video Transcript

Consider the series the sum from 𝑛 equals zero to ∞ of negative two to the 𝑛th power over 𝑛 times three to the power of 𝑛 plus one. Calculate the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛. Hence, determine whether the series converges or diverges.

Let’s look back at our original series. We know the general form is the sum of π‘Ž 𝑛. So, we’re going to let π‘Ž 𝑛 be equal to the bit inside the sum. It’s negative two to the power of 𝑛 over 𝑛 times three to the power of 𝑛 plus one. We should now be able to see that in order to calculate the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛, we need to work out π‘Ž sub 𝑛 plus one. To achieve this, we replace 𝑛 with 𝑛 plus one. So, the numerator becomes negative two to the power of 𝑛 plus one. And then, the denominator becomes 𝑛 plus one times three to the power of 𝑛 plus one plus one or three to the power of 𝑛 plus two.

We need to calculate the limit as 𝑛 approaches ∞ 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 𝑛 approaches ∞ of the absolute value of negative two to the power of 𝑛 plus one over 𝑛 plus one times three to the power of 𝑛 plus two divided by negative two to the 𝑛th power over 𝑛 times three to the power of 𝑛 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 𝑛 plus one over 𝑛 plus one times three to the power of 𝑛 plus two times 𝑛 times three to the power of 𝑛 plus one over negative two to the 𝑛th 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 𝑛 plus one over negative two to the power of 𝑛 is negative two to the power of 𝑛 plus one minus 𝑛, which is negative two to the power of one or simply negative two. Similarly, three to the power of 𝑛 plus one divided by three to the power of 𝑛 plus two is three to the power of 𝑛 plus one minus 𝑛 plus two, which is three to the power of negative one or one-third. So, we’re simply left with three on the denominator. We can now rewrite our limit in a number of ways. I’ve chosen to separate negative two-thirds and 𝑛 over 𝑛 plus one. So, we want to calculate the limit as 𝑛 approaches ∞ of the absolute value of negative two-thirds times 𝑛 over 𝑛 plus one.

One of the laws of limits says we can take out any constant factors. Now, here we’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, let’s find the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 over 𝑛 plus one. And if we were to apply direct substitution right now, we’d get ∞ over ∞, 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 𝑛 on our denominator. So, we’re going to divide both the numerator and the denominator by 𝑛. Let’s clear some space.

The fraction becomes 𝑛 over 𝑛 over 𝑛 over 𝑛 plus one over 𝑛. But of course, 𝑛 over 𝑛 is equal to one. And we need to calculate two-thirds of the limit as 𝑛 approaches ∞ of the absolute value of one over one plus one over 𝑛. Now, as 𝑛 grows larger, one over 𝑛 grows smaller. In other words, as 𝑛 approaches ∞, one over 𝑛 approaches zero. And so, the limit as 𝑛 approaches ∞ of the absolute value of one over one plus one over 𝑛 becomes the absolute value of one over one, which is simply one. Two-thirds times one is two-thirds. And so, we’ve calculated our limit. The limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛 is two-thirds.

So, how do we use this to establish whether the series converges or diverges? Well, let’s say the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž sub 𝑛 plus one over π‘Ž sub 𝑛 is equal to 𝐿, the constant 𝐿. The ratio test tells us that if 𝐿 is less than one, the series is absolutely convergent and hence converges. If it’s greater than one, the series diverges. And if it’s equal to one, our test is inconclusive. Well, our value for 𝐿 was two-thirds; this is quite clearly less than one. And so, by the ratio test, our series converges.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.