Video: Using the Comparison Test

Use the comparison test to decide whether the series βˆ‘_(𝑛 = 1)^(∞) 237/(𝑛 + 1.1^(𝑛)) is convergent or divergent.

03:32

Video Transcript

Use the comparison test to decide whether the series which is the sum from 𝑛 equals one to ∞ of 237 over 𝑛 plus 1.1 to the 𝑛 is convergent or divergent.

Let’s start by recalling the comparison test. The comparison test tells us that for the series which is the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛, and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where π‘Žπ‘› and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. That firstly if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum 𝑛 equals one to ∞ of π‘Ž 𝑛 also converges. And secondly, if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 diverges, and π‘Ž 𝑛 is greater or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 also diverges.

Now, our series has been given to us in the question, which gives that π‘Ž 𝑛 is equal to 237 over 𝑛 plus 1.1 to the 𝑛. Now, it looks as though we may be able to compare our series to a 𝑝 series. And this 𝑝 series is 237 multiplied by the sum from 𝑛 equals one to ∞ of one over 𝑛. Now, for this 𝑝 series, 𝑝 is equal to one. Therefore, this series diverges. Therefore, we’re looking at the second part of the comparison test. And we need to try and prove that π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛. So, that’s trying to prove that 237 over 𝑛 plus 1.1 to the 𝑛 is greater than or equal to 237 over 𝑛.

Now, the only differences in these two fractions is the denominators, in particular the term of 1.1 to the 𝑛. Since 𝑛 is always greater than or equal to one, therefore, 1.1 to the 𝑛 is always greater than zero. This tells us that the denominator of the fraction on the left will be greater than the denominator of the fraction at the right. And because of this, our inequality here does not hold, and so we cannot use this 𝑝 series to apply the comparison test. We’ll need to try a different series.

We can, in fact, try a geometric series. Geometric series is of the form of the sum of π‘Ž π‘Ÿ to the 𝑛. And a geometric series is convergent if the absolute value of π‘Ÿ is less than one and divergent if the absolute value of π‘Ÿ is greater than or equal to one. We now need to find a geometric series to compare our series to. Looking at our series, we have 1.1 to the 𝑛 in the denominator. Therefore, let’s try comparing our series to the sum from 𝑛 equals one to ∞ of 237 over 1.1 to the 𝑛.

For both of these series, since 𝑛 is greater than or equal to one, we have that both π‘Ž 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all 𝑛. Now, we can rewrite our geometric series in a form which is easier to find its convergence. And that is the sum from 𝑛 equals one to ∞ of 237 multiplied by one over 1.1 to the power of 𝑛. Here, we can see that π‘Ÿ is equal to one over 1.1. Since one over 1.1 is less than one, we have that this geometric series is convergent. Hence, we are looking at the first part of the comparison test.

We next need to try and prove the π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛. So, that is that 237 over 𝑛 plus 1.1 to the 𝑛 is less than or equal to 237 over 1.1 to the 𝑛. Since 𝑛 is greater than or equal to one, we have that the denominator of 𝑛 plus 1.1 to the 𝑛 is greater than the denominator of 1.1 to the 𝑛. And therefore, our inequality holds. Hence, we have satisfied the condition of the second part of the comparison test. And so, we reach our conclusion, which is that the sum from 𝑛 equals one to ∞ of 237 over 𝑛 plus 1.1 to the 𝑛 is convergent.

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